Question 15 Marks
Calculate mean deviation about median age distribution of $100$ persons given below:
|
Age
|
16-20
|
21-25
|
26-30
|
31-35
|
36-40
|
41-45
|
46-50
|
51-55
|
|
No. of persons
|
5
|
6
|
12
|
14
|
26
|
12
|
16
|
9
|
AnswerConverting the given data into continuous frequency distribution by subtrading $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval.
|
Age
|
$x_i$
|
$f_i$
|
Comulativefrequency
|
$|d_i| = |x_i - 38|$
|
$f_i|d_i|$
|
| 15.5-20.5 |
18 |
5 |
5 |
20 |
100 |
| 20.5-25.5 |
23 |
6 |
11 |
15 |
90 |
| 25.5-30.5 |
28 |
12 |
23 |
10 |
120 |
| 30.5-35.5 |
33 |
14 |
37 |
5 |
70 |
| 35.5-40.5 |
38 |
26 |
63 |
0 |
0 |
| 40.5-45.5 |
43 |
12 |
75 |
5 |
60 |
| 45.5-50.5 |
48 |
16 |
91 |
10 |
160 |
| 50.5-55.5 |
53 |
9 |
100 |
15 |
135 |
|
|
|
$\text{N}=\sum\text{f}_\text{i}=100$
|
|
|
$\sum\text{f}_\text{i}|\text{d}_\text{i}|=735$
|
clearly, N = 100 $\Rightarrow\frac{\text{N}}{2}=50.$ Cumulative frequency is just greater than $\frac{\text{N}}{2}$ is $63$ and the corresponding class is $35.5 - 40.5. l = 35.5, f = 26, h = 5, F = 37$
Therefore, $\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}=35.5+\frac{50-37}{26}\times5=38$
$\text{M.D}=\frac{1}{\text{N}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{735}{100}=7.35$ View full question & answer→Question 25 Marks
Find the mean deviation from the median for the following data:
|
$x_i$
|
$15$
|
$21$
|
$27$
|
$30$
|
|
$f_i$
|
$3$
|
$5$
|
$6$
|
$7$
|
Answer
|
$x_i$
|
$f_i$
|
Cum Freq
|
$|d_i| = |x_i - 30|$
|
$f_i|d_i|$
|
|
$15$
|
$3$
|
$3$
|
$15$
|
$45$
|
|
$21$
|
$5$
|
$8$
|
$9$
|
$45$
|
|
$27$
|
$6$
|
$14$
|
$3$
|
$18$
|
|
$30$
|
$7$
|
$21$
|
$0$
|
$0$
|
|
$35$
|
$8$
|
$29$
|
$5$
|
$40$
|
|
|
$29$
|
|
|
Total $= 148$
|
$\frac{\text{N}}{2}=14.5$
Median $= 30$
$\text{M.D}=\frac{148}{29}\approx5.10$ View full question & answer→Question 35 Marks
The number of telephone calls received at an exchange in 245 successive one- minute intervals are shown in the following frequency distribution:
|
Number of calls
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
|
Frequency
|
14
|
21
|
25
|
43
|
51
|
40
|
39
|
12
|
Compute the mean deviation about median.
AnswerWe have to calculate mean deviation from the median. So, first we calculate the median.
|
x
|
f
|
cf
|
d = (x-med)
|
fd
|
|
0
|
1
|
14
|
4
|
56
|
|
1
|
21
|
35
|
3
|
63
|
|
2
|
25
|
60
|
2
|
50
|
|
3
|
43
|
103
|
1
|
43
|
|
4
|
51
|
154
|
0
|
0
|
|
5
|
40
|
194
|
1
|
40
|
|
6
|
39
|
233
|
2
|
78
|
|
7
|
12
|
245
|
3
|
36
|
|
|
245
|
|
|
366
|
we have $\text{N}=245\Rightarrow\frac{\text{N}}{2}=122.5$ The cumulative frequency just greater than $\frac{\text{N}}{2}$ is 154 and the corresponding value of x is 4. Hence, median = 4 $\therefore\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{245}[366]=1.49$ View full question & answer→Question 45 Marks
Find the mean deviation from the mean for following data:
|
$x_i$
|
$5$
|
$7$
|
$9$
|
$10$
|
$12$
|
$15$
|
|
$f_i$
|
$8$
|
$6$
|
$2$
|
$2$
|
$2$
|
$6$
|
Answer
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
$|x_i -$ $\overline{\text{x}}$|
|
$f_i|x_i - 9|$
|
|
$5$
|
$8$
|
$40$
|
$4$
|
$32$
|
|
$7$
|
$6$
|
$42$
|
$2$
|
$12$
|
|
$9$
|
$2$
|
$18$
|
$0$
|
$0$
|
|
$10$
|
$2$
|
$20$
|
$1$
|
$2$
|
|
$12$
|
$2$
|
$24$
|
$3$
|
$6$
|
|
$15$
|
$6$
|
$90$
|
$6$
|
$36$
|
|
|
$\text{N}=\sum\text{f}_\text{i}=26$
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=234$
|
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-9|=88$
|
$\overline{\text{x}}=\frac{\sum_\limits{\text{i}=1}^\text{n}}{\text{N}}=\frac{234}{26}=9$
$\text{M.D}=\frac{1}{\text{N}}\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=\frac{1}{26}\times88=3.39$ View full question & answer→Question 55 Marks
Find the mean deviation from the mean for following data:
|
$x_i$
|
10
|
30
|
50
|
70
|
90
|
|
$f_i$
|
4
|
24
|
28
|
16
|
8
|
Answer
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
$|x_i - \overline{\text{x}}|$
|
$f_i|x_i - 14|$
|
|
10
|
4
|
40
|
40
|
160
|
|
30
|
24
|
720
|
20
|
480
|
|
50
|
28
|
1400
|
0
|
0
|
|
70
|
16
|
1120
|
20
|
320
|
|
90
|
8
|
720
|
40
|
320
|
|
|
N = 80
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=4000 $
|
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-50|=1280$
|
$\overline{\text{x}}=\frac{\sum_\limits{\text{i}=1}^\text{n}}{\text{N}}=\frac{4000}{80}=50$
$\text{M.D}=\frac{1}{\text{N}}\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=\frac{1}{80}\times1280=16$ View full question & answer→Question 65 Marks
Find the mean deviation from the median for the following data:
|
xi
|
74
|
89
|
42
|
54
|
91
|
94
|
35
|
|
fi
|
20
|
12
|
2
|
4
|
5
|
3
|
4
|
AnswerWe have to calculate mean deviation from the median. So, first we calculate the median.
|
x
|
f
|
Cf
|
d = (x-med)
|
fd
|
|
35
|
4
|
4
|
39
|
156
|
|
42
|
2
|
6
|
32
|
64
|
|
54
|
4
|
10
|
20
|
80
|
|
74
|
20
|
30
|
0
|
0
|
|
89
|
12
|
42
|
15
|
180
|
|
91
|
5
|
47
|
17
|
85
|
|
94
|
3
|
50
|
20
|
60
|
|
|
50
|
|
|
625
|
we have $\text{N}=50\Rightarrow\frac{\text{N}}{2}=25$ The cumulative frequency just greater than $\frac{\text{N}}2{}$ is 30 and the corresponding value of x is 74. Hence, median = 74 $\therefore\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{50}[625]=12.5$ View full question & answer→Question 75 Marks
Calculate the standard deviation for the following data:
|
Class:
|
0-30 |
30-60 |
60-90
|
90-120
|
120-150 |
150-180 |
180-210
|
| Frequency: |
9
|
17
|
43
|
82
|
81 |
44 |
24
|
Answer
| CI |
f |
x |
$\text{u}=\frac{(\text{x}-\text{A})}{\text{h}}$ |
f*u |
$u^2$ |
$fu^2$ |
| 0-30 |
9 |
15 |
-3 |
-27 |
9 |
81 |
| 30-60 |
17 |
45 |
-2 |
-34 |
4 |
68 |
| 60-90 |
43 |
75 |
-1 |
-43 |
1 |
43 |
| 90-120 |
82 |
105 |
0 |
0 |
0 |
0 |
| 120-150 |
81 |
135 |
1 |
81 |
1 |
81 |
| 150-180 |
44 |
165 |
2 |
88 |
4 |
176 |
| 180-210 |
24 |
195 |
3 |
72 |
9 |
216 |
| |
|
|
|
|
|
|
| |
90 |
|
|
10 |
|
150 |
Here, N = 300, A = 105, $\sum\text{f}_{\text{i}}\text{u}_{\text{i}}=137,\ \sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2=665$ and h = 30
$\therefore\text{Mean}=\overline{\text{x}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)$
$\Rightarrow\overline{\text{x}}=105+30\Big(\frac{137}{300}\Big)=118.7$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)^2\bigg]$
$\text{Var}(\text{X})=900\bigg[\frac{665}{300}-\Big(\frac{137}{300}\Big)^2\bigg]=1807.31$
$\therefore\text{SD}=\sqrt{\text{var}(\text{x})}=\sqrt{1807.31}=42.51$ View full question & answer→Question 85 Marks
Find the mean deviation from the mean for following data:
|
$x_i$
|
5
|
10
|
15
|
20
|
25
|
|
$f_i$
|
7
|
4
|
6
|
3
|
5
|
Answer
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
$|x_i - \overline{\text{x}}|$
|
$f_i|x_i - 14|$
|
|
5
|
7
|
35
|
9
|
63
|
|
10
|
4
|
40
|
4
|
16
|
|
15
|
6
|
90
|
1
|
6
|
|
20
|
3
|
60
|
6
|
18
|
|
25
|
5
|
125
|
11
|
55
|
|
|
N = 25
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=350$
|
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-14|=158$
|
$\overline{\text{x}}=\frac{\sum_\limits{\text{i}=1}^\text{n}}{\text{N}}=\frac{350}{25}=14$
$\text{M.D}=\frac{1}{\text{N}}\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=\frac{1}{25}\times158=6.32$ View full question & answer→Question 95 Marks
Find the mean deviation from the mean for following data:
|
Size
|
20
|
21
|
22
|
23
|
24
|
|
Freaquency
|
6
|
4
|
5
|
1
|
4
|
Answer
|
Size $(x_i)$
|
Frequency $(f_i)$
|
$f_ix_i$
|
$|x_i -\overline{\text{x}}| = |xi - 21.65|$
|
$f_i|x_i - \overline{\text{x}}| = fi|xi - 21.65|$
|
|
20
|
6
|
120
|
1.65
|
9.9
|
|
21
|
4
|
84
|
0.65
|
2.6
|
|
22
|
5
|
110
|
0.35
|
1.75
|
|
23
|
1
|
23
|
1.35
|
1.35
|
|
24
|
4
|
|
2.35
|
9.4
|
|
|
N = 20
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=433$
|
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=25$
|
$\overline{\text{x}}=\frac{\sum_\limits{\text{i}=1}^\text{n}}{\text{N}}=\frac{433}{20}=21.65$
$\text{MD}=\frac{1}{\text{N}}\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=\frac{1}{20}\times25=1.25$ View full question & answer→Question 105 Marks
calculate the mean deviation from the mean for the following data:
$36, 72, 46, 42, 60, 45, 53, 46, 51, 49$
AnswerLet ${\overline{\text{x}}}$ be the mean of the given data.
${\overline{\text{x}}}=\frac{36+72+46+42+60+45+53+46+51+59}{10}=50$
| $x_i$ |
$|d_i| = |x_i - {\overline{\text{x}}}|$ |
| 36 |
14 |
| 72 |
22 |
| 46 |
4 |
| 42 |
8 |
| 60 |
10 |
| 45 |
5 |
| 53 |
3 |
| 46 |
4 |
| 51 |
1 |
| 49 |
1 |
| Total |
72 |
we have,
$\sum|\text{x}_\text{i}-50|=\sum\text{d}_\text{i}=72$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{10}[72]=7.2$ View full question & answer→Question 115 Marks
Compute mean deviation from mean of the following distribution:
|
Marks
|
10-20
|
20-30
|
30-40
|
40-50
|
50-60
|
60-70
|
70-80
|
80-90
|
|
No. of students
|
8
|
10
|
15
|
25
|
20
|
18
|
9
|
5
|
AnswerComputation of mean deviation from the mean:
|
Marks
|
Number of students $f_i$
|
Midpoints $x_i$
|
$f_ix_i$
|
$|x_i - \overline{\text{X}}| |x_i - 49|$
|
$f_i|x_i - \overline{\text{X}}|$
|
|
10-20
|
8
|
15
|
120
|
34
|
272
|
|
20-30
|
10
|
25
|
250
|
24
|
240
|
|
30-40
|
15
|
35
|
525
|
14
|
210
|
|
40-50
|
25
|
45
|
1125
|
4
|
100
|
|
50-60
|
20
|
55
|
1100
|
6
|
120
|
|
60-70
|
18
|
65
|
1170
|
16
|
288
|
|
70-80
|
9
|
75
|
675
|
26
|
234
|
|
80-90
|
5
|
85
|
425
|
36
|
180
|
|
|
$\text{N}=\sum_\limits{\text{i}=1}^8\text{f}_\text{i}=110$
|
|
$\text{N}=\sum_\limits{\text{i}=1}^8\text{f}_\text{i}\text{x}_\text{i}=5390$
|
|
$\sum_\limits{\text{i}=1}^8\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{X}}=1644$
|
$\text{N}=\sum_\limits{\text{i}=1}^8\text{f}_\text{i}=110$
and $\sum_\limits{\text{i}=1}^8\text{f}_\text{i}\text{x}_\text{i}=5390$
$\overline{\text{X}}=\frac{\sum_\limits{\text{i}=1}^8\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{5390}{110}$
$= 49$
$\text{Mean}\ \text{deviation}=\frac{\sum_\limits{\text{i}=1}^8\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{X}}|}{\text{N}}$
$=\frac{1644}{110}$
$=14.945$
$\approx14.95$ View full question & answer→Question 125 Marks
An analysis of the weekly wages paid to workers in two firms $A$ and $B,$ belonging to the same industry gives the following results:
|
|
Firm $A$
|
Firm $B$
|
|
No. of wage earners
|
$586$ |
$648$ |
|
Average weekly wages
|
$52.5$ |
$47.5$ |
|
Variance of the
|
$100$ |
$121$ |
|
Distribution of wages
|
|
|
- Which firm $A$ or $B$ pays out larger amount as weekly wages?
- Which firm $A$ or $B$ has greater variability in individual wages?
AnswerTotal wagas paid by firm $A = ($Averge wages$) \times ($Number of employees$)$
$= 52.5 \times 587$
$= Rs. 30817.50$
Total wagas paid by firm $A = ($Averge wages$) \times ($Number of employees$)$
$= 47.5 \times 648$
$= Rs. 30780$
So, firm $A$ pays higher total wages.
In order to compare the variability of wages among the two firm, we have to calculate their coefficients of variation.
Let $\sigma_1$ and $\sigma_2$ denote the standard deviations of firm $A$ and firm $B$ respectively. Further,
Let $\overline{\text{X}}_1$ and $\overline{\text{X}}_2$ be the mean wages in firms $A$ and $B$ respectively.
We have,
$\overline{\text{X}}_1=52.5,\ \overline{\text{X}}_2=47.5$
$\sigma_1^2=100$ and $\sigma_2^2=121$
$\Rightarrow\sigma_1=\sqrt{100}=10$ and $\sigma_2=\sqrt{121}=11$
Now,
Coefficient of variation in wages in firm $\text{A}=\frac{\sigma_1}{\overline{\text{x}_1}}\times100$
$=\frac{10}{52.5}\times100$
$=19.05$
and,
Coefficient of variation in wages in firm $\text{B}=\frac{\sigma_2}{\overline{\text{x}_2}}\times100$
$=\frac{11}{47.5}\times100$
$=23.16$
Clearly, coefficient of variation in wages is greater for firm $B$ than for firm $A.$
So, firm $B$ shows more variability in wages.
View full question & answer→Question 135 Marks
Calculate the mean deviation of the following income groups of five and seven members from their medians:
|
I
Income in ₹
|
II
Income in ₹
|
|
4000
|
3800
|
| 4200 |
4000 |
| 4400 |
4200 |
| 4600 |
4400 |
| 4800 |
4600 |
| |
4800 |
| |
5800 |
AnswerArrange the given data for income group I in assending order, middle observation is 4400.
So, median = 4400.
Mean deviation for group I
|
$x_i$
|
$|d_i| = |x_i - 4400|$
|
|
4000
|
400
|
|
4200
|
200
|
|
4400
|
0
|
|
4600
|
200
|
|
4800
|
400
|
|
Total
|
$\sum$|di| = 1000
|
$\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1000}{5}=200$
Arrange the given data for income group II in assending order, middle observation is $4400$.
So, median = $4400$.
Mean deviation for group II
|
$x_i$
|
$|d_i| = |x_i - 4400|$
|
|
3800
|
600
|
|
4000
|
400
|
|
4200
|
200
|
|
4400
|
0
|
|
4600
|
200
|
|
4800
|
400
|
|
5800
|
1400
|
|
Total
|
$\sum$|di| = 3200
|
$\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{3200}{7}=457.14$ View full question & answer→Question 145 Marks
Find the mean deviation from the mean and from median of the following distribution:
|
Marks
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
|
No. of students
|
5
|
8
|
15
|
16
|
6
|
AnswerM.D from median
|
Marks
|
Students
|
$x_i$
|
Cum. Freq
|
$|\text{d}_\text{i}|=\Big|\text{x}_\text{i}-\frac{70}{3}\Big|$
|
$f_id_i$
|
|
0-10
|
5
|
5
|
5
|
$\frac{55}{3}$
|
$\frac{275}{3}$
|
|
10-20
|
8
|
15
|
13
|
$\frac{25}{3}$
|
$\frac{200}{3}$
|
|
20-30
|
15
|
25
|
28
|
$\frac{5}{3}$
|
$\frac{75}{3}$
|
|
30-40
|
16
|
35
|
44
|
$\frac{35}{3}$
|
$\frac{560}{3}$
|
|
40-50
|
6
|
45
|
50
|
$\frac{65}{3}$
|
$\frac{390}{3}$
|
|
|
N = 50
|
|
|
|
Total = 500
|
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=20+\frac{30-25}{15}\times10$
$=20+\frac{10}{3}=\frac{70}{3}$
$\text{M.D}=\frac{500}{50}=10$
M.D from mean
|
Marks
|
Students
|
$x_i$
|
$\text{d}_\text{i}=\frac{\text{x}_\text{i}-35}{10}$
|
$f_id_i$
|
$|x_i - 27|$
|
$f_i|x_i- 27|$
|
|
0-10
|
5
|
5
|
-3
|
-15
|
22
|
110
|
|
10-20
|
8
|
15
|
-2
|
-16
|
12
|
96
|
|
20-30
|
15
|
25
|
-1
|
-15
|
2
|
30
|
|
30-40
|
16
|
35
|
0
|
0
|
8
|
128
|
|
40-50
|
6
|
45
|
1
|
6
|
18
|
108
|
|
|
N = 50
|
|
|
Total = 40
|
|
Total = 472
|
$\overline{\text{X}}=35+10\times\frac{-40}{50}=27$
$\text{M.D}=\frac{472}{50}=9.44$ View full question & answer→Question 155 Marks
Find the numberof observation lying between $\overline{\text{X}}-\text{M.D. }$ and $\overline{\text{X}} +\text{ M.D.}$ is the mean deviation from the mean.
$38, 70, 48, 34, 63, 42, 55, 44, 53, 47$
AnswerLet $\overline{\text{x}}$ be the mean of the data set.
$\overline{\text{x}}=\frac{38+70+48+34+63+42+55+44+53+47}{10}=49.4$
|
$x_i$
|
$|d_i| = |x_i - 49.4|$
|
|
38
|
11.4
|
|
70
|
20.6
|
|
48
|
1.4
|
|
34
|
15.4
|
|
63
|
13.6
|
|
42
|
7.4
|
|
55
|
5.6
|
|
44
|
5.4
|
|
53
|
3.6
|
|
47
|
2.4
|
|
Total
|
86.8
|
$\text{MD}=\frac{1}{10}\times86.8=8.68$
$\overline{\text{x}}$ - M.D. = 49.4 - 8.68 = 40.72
and, $\overline{\text{x}}$ + M.D. = 49.4 + 8.68 =58.08
There are 6 observation between 40.72 and 58.08. View full question & answer→Question 165 Marks
Compute the mean deviation from the median of the following distribution:
|
Class
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
|
Frequency
|
5
|
10
|
20
|
5
|
10
|
AnswerWe have to calculate mean deviation from the median. So, first we calculate the median.
|
CI
|
x
|
f
|
cf
|
d = (x-med)
|
fd
|
|
0-10
|
5
|
5
|
5
|
20
|
100
|
|
10-20
|
15
|
10
|
15
|
10
|
100
|
|
20-30
|
25
|
20
|
35
|
0
|
0
|
|
30-40
|
35
|
5
|
91
|
10
|
50
|
|
40-50
|
45
|
10
|
101
|
20
|
200
|
|
|
|
50
|
|
|
450
|
$\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{50}[450]=9$ View full question & answer→Question 175 Marks
calculate the mean deviation from the median of the following frequency distribution:
|
Hights in inches
|
58
|
59
|
60
|
61
|
62
|
63
|
64
|
65
|
66
|
|
No. of students
|
15
|
20
|
32
|
35
|
35
|
22
|
20
|
10
|
8
|
Answer
|
$x_i$
|
$f_i$
|
Cum.Freq
|
$|d_i| = |x_i - 61|$
|
$f_i|d_i|$
|
|
58
|
15
|
15
|
3
|
45
|
|
59
|
20
|
35
|
2
|
40
|
|
60
|
32
|
67
|
1
|
32
|
|
61
|
35
|
102
|
0
|
0
|
|
62
|
35
|
137
|
1
|
35
|
|
63
|
22
|
159
|
2
|
44
|
|
64
|
20
|
179
|
3
|
60
|
|
65
|
10
|
189
|
4
|
40
|
|
66
|
8
|
197
|
5
|
40
|
|
|
N = 197
|
|
|
Total = 336
|
$\text{N}=197,\frac{\text{N}}{2}=98.5$
Corresponding value for median is 61
$\text{Mean}\ \text{Deviation}=\frac{336}{197}=1.705$ View full question & answer→Question 185 Marks
Calculate the mean deviation about the median of the following observation:
$34, 66, 30, 38, 44, 50, 40, 60, 42, 51$
AnswerFormula used for mean deviation:
$\text{MD}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
Here, $n = 10$
Also, Median is the AM of the fifth and the sixth observation.
Median, $\text{M}=\frac{42+484}{2}=43$
|
$x_i$
|
$|d_i| = |x_i- M|$
|
|
34
|
9
|
|
66
|
23
|
|
30
|
13
|
|
38
|
5
|
|
44
|
1
|
|
50
|
7
|
|
40
|
3
|
|
60
|
17
|
| 42 |
1 |
| 51 |
8 |
| Total |
87 |
$\text{MD}=\frac{1}{10}\times87=8.7$ View full question & answer→Question 195 Marks
Calculate the mean deviation about the median of the following observation:
$38, 70, 48, 34, 42, 55, 63, 46, 54, 44$
AnswerFormula used for mean deviation:
$\text{MD}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
Here, $n$ is equal to $10$.
Median is the arithmetic mean of the fifth and the sixth observation.
Median, $\text{M}=\frac{46+48}{2}=47$
|
$x_i$
|
$|d_i| = |x_i- M|$
|
|
38
|
9
|
|
70
|
23
|
|
48
|
1
|
|
34
|
13
|
|
42
|
5
|
|
55
|
8
|
|
63
|
16
|
|
46
|
1
|
| 54 |
7 |
| 44 |
3 |
| Total |
86 |
$\text{MD}=\frac{1}{10}\times86=8.6$ View full question & answer→Question 205 Marks
Find the mean deviation from the mean for the following data:
|
classes
|
95-105
|
105-115
|
115-125
|
125-135
|
135-145
|
145-155
|
|
Frequencies
|
9
|
13
|
16
|
26
|
30
|
12
|
Answer
|
Classes
|
$f_i$
|
$x_i$
|
$d_i$
|
$f_id_i$
|
$|x_i - \overline{\text{X}}|$
|
$f_i|x_i - \overline{\text{X}}|$
|
|
95-105
|
9
|
100
|
-3
|
-27
|
28.58
|
257.22
|
|
105-115
|
13
|
110
|
-2
|
-26
|
18.58
|
241.54
|
|
115-125
|
16
|
120
|
-1
|
-16
|
8.58
|
137.28
|
|
125-135
|
26
|
130
|
0
|
0
|
1.42
|
36.92
|
|
135-145
|
30
|
140
|
1
|
30
|
11.42
|
342.6
|
|
145-155
|
12
|
150
|
2
|
24
|
21.42
|
257.04
|
|
|
N = 106
|
|
|
Total = -15
|
|
Total = 1272.60
|
$N = 106$
$a = 130$
$h = 10$
$\overline{\text{X}}=\text{a+h}\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}\Big)=128.58$
$\text{M.D}=\frac{\sum\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{X}}}{\text{N}}=\frac{1272.60}{106}=12.005$ View full question & answer→Question 215 Marks
Calculate the mean deviation from the mean for the following data:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer$\text{Mean}=\frac{1}{\text{n}}\sum|\text{x}_\text{i}|=\frac{500}{10}=50$
Calculation of Mean Deviation
|
X-values
|
Deviation From Mean
|
|
38
|
12
|
|
70
|
20
|
| 48 |
2 |
| 40 |
10 |
|
42
|
8
|
| 55 |
5 |
|
63
|
13
|
|
46
|
4
|
|
54
|
4
|
|
44
|
6
|
|
Total
|
84
|
We have,
$\sum|\text{x}_\text{i}-50|=\sum\text{d}_\text{i}=84$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{10}[84]=8.4$ View full question & answer→Question 225 Marks
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
|
Subject
|
Mathematics
|
Physics
|
Chemistry
|
|
Mean
|
42
|
32
|
40.9
|
|
Standard
|
12
|
15
|
20
|
|
Deviation
|
|
|
|
Which of the three subjects shows the highest variability in marks and which shows the lowest?
AnswerIn order to compare the variability of mark in maths, Physics and Chemistry, we have to calculate their coefficients of variation.
Let $\sigma_1,\ \sigma_2$ and $\sigma_3$ denote the standard deviations of marks in Maths, Physics and Chemistry respectively. Further, Let $\overline{\text{X}}_1,\ \overline{\text{X}}_2$ and $\overline{\text{X}}_3$ be the mean score in Maths, Physics and Chemistry respectively.
We have,
$\overline{\text{X}}_1=42,\ \overline{\text{X}}_2=32,\ \overline{\text{X}}_3=40.9$
$\Rightarrow\sigma_1=12,\ \sigma_2=15,\ \sigma_3=20$
Now,
Coefficient of variation in Maths $=\frac{\sigma_1}{\overline{\text{x}_1}}\times100=\frac{12}{42}\times100=28.57$
Coefficient of variation in Physics $=\frac{\sigma_2}{\overline{\text{x}_2}}\times100=\frac{15}{32}\times100=46.88$
Coefficient of variation in Chemistry $=\frac{\sigma_3}{\overline{\text{x}_3}}\times100=\frac{20}{40.9}\times100=48.90$
Clearly, coefficient of variation in marks is greater in Chemistry and lowest in Maths.
So, marks in Chemistry show highest variability and marks in maths show lowest variability.
View full question & answer→Question 235 Marks
Calculate the mean, median and standard deviation of the following distribution:
| Class-interval: |
31-35 |
36-40 |
41-45 |
46-50 |
51-55 |
56-60 |
61-65 |
66-70 |
| Frequency: |
2 |
3 |
8 |
12 |
16 |
5 |
2 |
3 |
Answer
| Class Interval |
$f_i$ |
Midpoint $x_i$ |
$\text{u}_{\text{i}}=\frac{\text{x}_{\text{i}}-53}{4}$ |
$u_i^2$ |
$f_iu_i$ |
$f_iu_i^2$ |
| 31-35 |
2 |
33 |
-5 |
25 |
-10 |
50 |
| 36-40 |
3 |
38 |
-3.75 |
14.06 |
-11.25 |
42.18 |
| 41-45 |
8 |
43 |
-2.5 |
6.25 |
-20 |
50 |
| 46-50 |
12 |
48 |
-1.25 |
1.56 |
-15 |
18.72 |
| 51-55 |
16 |
53 |
0 |
0 |
0 |
0 |
| 56-60 |
5 |
58 |
1.25 |
1.56 |
6.25 |
7.8 |
| 61-65 |
2 |
63 |
2.5 |
6.25 |
5 |
12.5 |
| 66-70 |
3 |
68 |
3.75 |
14.06 |
11.25 |
42.18 |
| |
N = 51 |
|
|
|
$\sum^\limits{\text{n}}_{\text{i}=1}\text{f}_\text{i}\text{u}_\text{i}=-33.75$ |
$\sum^\limits{\text{n}}_{\text{i}=1}\text{f}_\text{i}\text{u}_\text{i}^2=223.38$ |
$\overline{\text{X}}=\text{a+h}\Bigg(\frac{\sum\limits^\text{n}_{\text{i}=1}\text{f}_{\text{i}}\text{u}_\text{i}}{\text{n}}\Bigg)$
$=53+4\Big(\frac{-33.75}{51}\Big)$
$=50.36$
$\sigma^2=\text{h}^2\Bigg(\frac{\sum\limits^\text{n}_{\text{i}=1}\text{f}_{\text{i}}\text{u}_\text{i}^2}{\text{n}}-\Bigg(\frac{\sum\limits^\text{n}_{\text{i}=1}\text{f}_{\text{i}}\text{u}_\text{i}}{\text{n}}\Bigg)^2\Bigg)$
$=16\Big(\frac{223.38}{51}-\frac{1139.06}{2601}\Big)$
$=63.07$
$\sigma=\sqrt{63.07}$
$=7.94$
|
$f_i$
|
CF (Cumulative frequency)
|
|
2
|
2
|
|
3
|
5
|
|
8
|
13
|
|
12
|
25
|
|
16
|
41
|
|
5
|
46
|
|
2
|
48
|
|
3
|
51
|
$\sum\text{f}_{\text{i}}=51=\text{N}$
$\frac{\text{N}}{2}=25.5$
Median class interval is $51-55$.
$L = 51$
$F = 25$
$f = 16$
$h = 4$
Median $=\text{L}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=51+\frac{25.5-25}{16}\times4$
$=51+\frac{0.5}{4}$
$=51.125$ View full question & answer→Question 245 Marks
Find the mean deviation from the mean for the following data:
|
classes
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
50-60
|
|
Frequencies
|
6
|
8
|
14
|
16
|
4
|
2
|
Answer
|
CI
|
x
|
f
|
xf
|
d = (x-mean)
|
fd
|
|
0-10
|
5
|
6
|
30
|
22
|
132
|
|
10-20
|
15
|
8
|
120
|
12
|
96
|
|
20-30
|
24
|
14
|
350
|
2
|
28
|
|
30-40
|
35
|
16
|
560
|
8
|
128
|
|
40-50
|
45
|
4
|
180
|
18
|
72
|
|
50-60
|
55
|
2
|
110
|
28
|
56
|
|
|
|
50
|
1350
|
|
512
|
|
|
Mean
|
|
27
|
|
|
|
|
Mean Deviation
|
|
10.24
|
|
|
$\text{Mean}=\frac{1}{\text{n}}\sum\text{f}_\text{i}\text{x}_\text{i}=\frac{1350}{50}=27$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{50}[512]=10.24$ View full question & answer→Question 255 Marks
Find the mean deviation from the mean for the following data:
|
Classes
|
0-100
|
100-200
|
200-300
|
300-400
|
400-500
|
500-600
|
600-700
|
700-800
|
|
Frequencies
|
4
|
8
|
9
|
10
|
7
|
5
|
4
|
3
|
Answer
|
CI
|
x
|
f
|
xf
|
d = (x-mean)
|
fd
|
|
0-100
|
50
|
4
|
200
|
308
|
1232
|
|
100-200
|
150
|
8
|
1200
|
208
|
1664
|
|
200-300
|
250
|
9
|
2250
|
108
|
972
|
|
300-400
|
350
|
10
|
3500
|
8
|
80
|
|
400-500
|
450
|
7
|
3150
|
92
|
644
|
|
500-600
|
550
|
5
|
2750
|
192
|
960
|
|
600-700
|
650
|
4
|
2600
|
292
|
1168
|
|
700-800
|
750
|
3
|
2250
|
392
|
1176
|
|
|
|
50
|
17900
|
|
7896
|
$\text{Mean}=\frac{1}{\text{n}}\sum\text{f}_\text{i}\text{x}_\text{i}=\frac{17900}{50}=358$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{50}[7896]=157.92$ View full question & answer→Question 265 Marks
Calculate coefficient of variation from the following data:
|
Income (in Rs):
|
1000-1700
|
1700-2400
|
2400-3100
|
3100-3800
|
3800-4500
|
4500-5
|
|
No. of families:
|
12
|
18
|
20
|
25
|
35
|
10
|
Answer
|
$CI$
|
$f$
|
$x$
|
$\text{u}=\frac{(\text{x}-\text{A})}{\text{h}}$
|
$fu$
|
$`u^2$
|
$fu^2$
|
|
1000-1700
|
12
|
1350
|
-2
|
-24
|
4
|
48
|
|
1700-2400
|
18
|
2050
|
-1
|
-18
|
1
|
18
|
|
2400-3100
|
20
|
2750
|
0
|
0
|
0
|
0
|
|
3100-3800
|
25
|
3450
|
1
|
25
|
1
|
25
|
|
3800-4500
|
35
|
4150
|
2
|
70
|
4
|
140
|
|
4500-5200
|
10
|
4850
|
3
|
30
|
9
|
90
|
|
|
120
|
|
|
83
|
|
321
|
Here, N = 120, A = 2750, $\sum\text{f}_\text{i}\text{u}_\text{i}=83,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=321$ and h = 700
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=2750+700\Big(\frac{83}{120}\Big)=3234.17$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=490000\bigg[\frac{321}{120}-\Big(\frac{83}{120}\Big)^2\bigg]=1076332.64$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{1076332.64}=1037.46$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}_1}\times100=\frac{1037.46}{3234.17}\times100=32.08$ View full question & answer→Question 275 Marks
Calculate the mean deviation about mean for the following distribution:
|
Class interval
|
0-4
|
4-8
|
8-12
|
12-16
|
16-20
|
|
Frequency
|
4
|
6
|
8
|
5
|
2
|
Answer
|
Classes
|
$f_i$
|
$x_i$
|
$f_ix_i$
|
$|x_i- 9.2|$
|
$f_i|x_i - 9.2|$
|
|
0-4
|
4
|
2
|
8
|
7.2
|
28.8
|
|
4-8
|
6
|
6
|
36
|
3.2
|
19.2
|
|
8-12
|
8
|
10
|
80
|
0.8
|
6.4
|
|
12-16
|
5
|
14
|
70
|
4.8
|
24.0
|
|
16-20
|
2
|
18
|
36
|
8.8
|
17.6
|
|
|
N = 25
|
|
Total = 230
|
|
Total = 96.0
|
$\text{Mean}=\frac{230}{25}=9.2$
$\text{M.D}=\frac{96}{25}=3.84$ View full question & answer→Question 285 Marks
The age distribution of 100 life-insuance policy holders is an follows:
|
Age (on nearest birth day)
|
17-19.5
|
20-25.5
|
26-35.5
|
36-40.5
|
41-50.5
|
51-55.5
|
56-60.5
|
61-70.5
|
|
No. of persons
|
5
|
16
|
12
|
26
|
14
|
12
|
6
|
5
|
AnswerWe have to calculate mean deviation from the median. So, first we calculate the median.
|
CI
|
x
|
f
|
cf
|
d = (x-med)
|
fd
|
|
17-19.5
|
18.25
|
5
|
5
|
20
|
100
|
|
20-25.5
|
22.75
|
16
|
21
|
15.5
|
248
|
|
26-35.5
|
30.75
|
12
|
33
|
7.5
|
90
|
|
36-40.5
|
38.25
|
26
|
59
|
0
|
0
|
|
41-50.5
|
45.75
|
14
|
73
|
7.5
|
105
|
|
51-55.5
|
53.25
|
12
|
85
|
15
|
180
|
|
56-60.5
|
58.25
|
6
|
91
|
20
|
120
|
|
61-70.5
|
65.75
|
5
|
96
|
27.5
|
137.5
|
|
|
|
96
|
|
|
980.5
|
we have N = 96 ⇒ $\frac{\text{N}}{2}=48$ the cumulative frequency just greater than $\frac{\text{N}}{2}$ is 59 and the corresponding value of x is 38.25. Hence, median = 38.25$\therefore\text{M.D}=\frac{1}{\text{n}}\sum\text{f}_\text{i}|\text{d}_\text{i}|=\frac{1}{96}[980.5]=10.21$ View full question & answer→Question 295 Marks
Calculate the mean and S.D. for the following data:
|
Expenditere(in ₹):
|
0-10 |
10-20 |
20-30
|
30-40
|
40-50
|
| Frequency: |
14
|
13
|
27
|
21
|
15
|
Answer
| $CI$ |
$f$ |
$x$ |
$\text{u}=\frac{(\text{x}-\text{A})}{\text{h}}$ |
$fu$ |
$u^2$ |
$fu^2$ |
| 0-10 |
14 |
5 |
-2 |
-28 |
4 |
56 |
| 10-20 |
13 |
15 |
-1 |
-13 |
1 |
13 |
| 20-30 |
27 |
25 |
0 |
0 |
0 |
0 |
| 30-40 |
21 |
35 |
1 |
21 |
1 |
21 |
| 40-50 |
15 |
45 |
2 |
30 |
4 |
60 |
| |
90 |
|
|
10 |
|
150 |
Here, N = 90, A = 25, $\sum\text{f}_{\text{i}}\text{u}_{\text{i}}=10,\ \sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2=150$ and h = 10
$\therefore\text{Mean}=\overline{\text{x}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)$
$\Rightarrow\overline{\text{x}}=25+10\Big(\frac{10}{90}\Big)=26.11$
$\text{Var}(\text{X})=\text{h}^2\Big[\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)^2\Big]$
$\text{Var}(\text{X})=100\Big[\frac{150}{90}-\Big(\frac{10}{90}\Big)^2\Big]=165.4$
$\therefore\text{SD}=\sqrt{\text{var}(\text{x})}=\sqrt{165.4}=12.86$ View full question & answer→Question 305 Marks
Calculate the mean deviation about the median of the following observation:
$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$
AnswerFormula used for mean deviation:
$\text{MD}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
Here, $n = 10$
Also, Median is the AM of the fifth and the sixth observation.
Median, $\text{M}=\frac{28+29}{2}=28.5 $
| $X_i$ |
$|d_i| = |x_i - M|$ |
| 22 |
6.5 |
| 24 |
4.5 |
| 30 |
1.5 |
| 27 |
1.5 |
| 29 |
0.5 |
| 31 |
2.5 |
| 25 |
3.5 |
| 28 |
0.5 |
| 41 |
12.5 |
| 41 |
13.5 |
| Total |
47 |
$\text{MD}=\frac{1}{10}\times47=4.7$ View full question & answer→Question 315 Marks
The variance of 20 observation is 5. If each observation is multiplied by 2, find the variance of the resulting observation.
AnswerWe have, n = 20, and $\sigma^2=5$
Now each observation is multiplied by 2.
Suposs X = 2x be the new data.
$\therefore\overline{\text{X}}=\frac{1}{20}\sum2\text{x}_\text{i}=\frac{1}{20}\times2\sum\text{x}_\text{i}=2\overline{\text{x}}$
$\Rightarrow\sum{\text{X}_\text{i}}^{2}=4\sum{\text{x}_\text{i}}^{2}$
Since, $\sigma^2=5$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=5$
Now, for the new data
$\sigma^2=\frac{1}{\text{n}}\sum{\text{X}_\text{i}}^2-(\overline{\text{x}})^2=4\sum{\text{x}_\text{i}}^2-(2\overline{\text{x}})^2=4\Big(\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big)=4\times5=20$
View full question & answer→Question 325 Marks
Find the mean deviation from the mean for following data:
|
Size
|
1
|
3
|
5
|
7
|
9
|
11
|
13
|
15
|
|
Frequency
|
3
|
3
|
4
|
14
|
7
|
4
|
3
|
4
|
Answer
|
Size $(x_i)$
|
Frequency $(f_i)$
|
$f_ix_i$
|
$|x_i - \overline{\text{x}}| = |xi - 8|$
|
$f_i|x_i - \overline{\text{x}}| = fi|xi - 8|$
|
|
1
|
3
|
3
|
7
|
21
|
|
3
|
3
|
9
|
5
|
15
|
|
5
|
4
|
20
|
3
|
12
|
|
7
|
14
|
98
|
1
|
14
|
|
9
|
7
|
63
|
1
|
7
|
|
11
|
4
|
44
|
3
|
12
|
|
13
|
3
|
39
|
5
|
15
|
|
15
|
4
|
60
|
7
|
28
|
|
|
N = 42
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=336$
|
|
$\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=124$
|
$\overline{\text{x}}=\frac{\sum_\limits{\text{i}=1}^\text{n}}{\text{N}}=\frac{336}{42}=8$
$\text{MD}=\frac{1}{\text{N}}\sum_\limits{\text{i}=1}^\text{n}\text{f}_\text{i}|\text{x}_\text{i}-\overline{\text{x}}|=\frac{1}{42}\times124=2.95$ View full question & answer→Question 335 Marks
calculate the mean deviation from the mean for the following data:
$57, 64, 43, 67, 49, 59, 44, 47, 61, 59$
AnswerFirst arrange the given numbers in assending order
write these numbers in assending order
$57, 64, 43, 67, 49, 59, 44, 47, 61, 59$
we get $43, 44, 47, 49, 57, 59, 59, 61, 64, 67$
Let X be the mean of given data, we get
$\text{X}=\frac{43+44+47+49+57+59+59+61+64+67}{10}=55$
Calculationof Mean Deviations from mean
| $x_i$ |
$|d_i| = |x_i - 55|$ |
| 43 |
12 |
| 44 |
11 |
| 47 |
8 |
| 49 |
6 |
| 57 |
2 |
| 59 |
4 |
| 59 |
4 |
| 61 |
6 |
| 64 |
9 |
| 67 |
12 |
| Total |
74 |
$\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}$
$=\frac{74}{10}$
$=7.4$
View full question & answer→Question 345 Marks
Calculate the mean deviation about the median of the following observation:
$38, 70, 48, 34, 63, 42, 55, 44, 53, 47$
AnswerFormula used for mean deviation:
$\text{MD}=\frac{1}{\text{n}}\sum_\limits{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
Here,
$d_i = x_i - M$
M = median
Here, $n = 10$.
also, median is the AM of the fifth and sixth observation.
Median, $\text{M}=\frac{47+48}{2}=47.5$
| $X_i$ |
$|d_i| = |x_i - M|$ |
| 38 |
9.5 |
| 70 |
22.5 |
| 48 |
0.5 |
| 34 |
13.5 |
| 63 |
15.5 |
| 42 |
5.5 |
| 55 |
7.5 |
| 44 |
3.5 |
| 47 |
0.5 |
| Total |
84 |
$\text{MD}=\frac{1}{10}\times84=8.4$ View full question & answer→Question 355 Marks
Calculate the mean deviation about the median of the following observation:
$3011, 2780, 3020, 2354, 3541, 4150, 5000$
AnswerFormula used for mean deviation:
$\text{MD}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
Here,
$d_i = x_i - M$
m = Median
Here, median (M) = 3020 and $n = 7$.
| $X_i$ |
$|d_i| = |x_i- 3020|$
|
|
3011
|
9
|
|
2780
|
240
|
|
3020
|
0
|
|
2354
|
666
|
|
3541
|
521
|
|
4150
|
1130
|
|
5000
|
1980
|
|
Total
|
4546
|
$\text{MD}=\frac{1}{\text{n}}\sum\limits_{\text{i}=1}^\text{n}|\text{d}_\text{i}|$
$\text{MD}=\frac{1}{7}\times4546$
$=649.42$ View full question & answer→Question 365 Marks
Find the mean variance and standard deviation for the following data:
$227, 235, 255, 269, 292, 299, 312, 321, 333, 348.$
Answer
|
$x_i$
|
$d_i = x_i - 299$
|
$d_i{}^2$
|
|
227
|
-72
|
5184
|
|
235
|
-64
|
4096
|
|
255
|
-44
|
1936
|
|
269
|
-30
|
900
|
|
292
|
-7
|
49
|
|
299
|
0
|
0
|
|
312
|
13
|
169
|
|
321
|
22
|
484
|
|
333
|
34
|
1156
|
|
348
|
49
|
2401
|
|
|
Total = -99
|
Total = 16375
|
$\overline{\text{X}}=299+\frac{-99}{10}=289.1$
$\text{var}=\frac{16375}{10}-\Big(\frac{-99}{10}\Big)^2=1637.5-98.01=1539.49$
$\text{S.D}=\sqrt{1539.49}=39.24$ View full question & answer→Question 375 Marks
Find the mean variance and standard deviation for the following data:
6, 7, 10, 12, 13, 4, 8, 12.
Answer
|
x
|
d = (x - Mean)
|
$d^2$
|
|
6
|
-3
|
9
|
|
7
|
-2
|
4
|
|
10
|
1
|
1
|
|
12
|
3
|
9
|
|
13
|
4
|
16
|
|
4
|
-5
|
25
|
|
8
|
-1
|
1
|
|
12
|
3
|
9
|
|
72
|
|
74
|
$\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}=\frac{1}{8}[72]=9$
$\text{var}(\text{x})=\frac{1}{\text{n}}\Big\{\sum(\text{x}_\text{i}-\overline{\text{x}})^2\Big\}=\frac{1}{8}\big\{74\big\}=9.25$
$\text{S.D}(\text{x})=\sqrt{\text{var}(\text{x})}=\sqrt{9.25}=3.04$ View full question & answer→Question 385 Marks
Calculate the mean deviation from the median of the following data:
|
Class interval
|
0-6
|
6-12
|
12-18
|
18-24
|
24-30
|
|
Frequency
|
4
|
5
|
3
|
6
|
2
|
Answer
|
Classes
|
$f_i$
|
$x_i$
|
$f_ix_i$
|
$|x_i- 14.1|$
|
$f_i|x_i - 14.1|$
|
|
0-6
|
4
|
3
|
12
|
11.1
|
44.4
|
|
6-12
|
5
|
9
|
45
|
5.1
|
25.5
|
|
12-18
|
3
|
15
|
45
|
0.9
|
2.7
|
|
18-24
|
6
|
21
|
126
|
6.9
|
41.4
|
|
24-30
|
2
|
27
|
54
|
12.9
|
25.8
|
|
|
N = 20
|
|
Total = 282
|
|
Total = 139.8
|
$\text{Mean}=\frac{282}{20}=14.1$
$\text{M.D}=\frac{139}{20}=6.99$ View full question & answer→Question 395 Marks
Calculate the $A.M$. and $S.D$. for the following distribution:
|
Class:
|
0-10 |
10-20 |
20-30
|
30-40
|
40-50 |
50-60 |
60-70 |
70-80
|
| Frequency: |
18
|
16
|
15
|
12
|
10 |
5 |
2 |
1
|
Answer
| $CI$ |
$f$ |
$x$ |
$\text{u}=\frac{(\text{x}-\text{A})}{\text{h}}$ |
$f*u$ |
$u^2$ |
$fu^2$ |
| 0-10 |
18 |
5 |
-3 |
-54 |
9 |
162 |
| 10-20 |
16 |
15 |
-2 |
-32 |
4 |
64 |
| 20-30 |
15 |
25 |
-1 |
-15 |
1 |
15 |
| 30-40 |
12 |
35 |
0 |
0 |
0 |
0 |
| 40-50 |
10 |
45 |
1 |
10 |
1 |
10 |
| 50-60 |
5 |
55 |
2 |
10 |
4 |
20 |
| 60-70 |
2 |
65 |
3 |
6 |
9 |
18 |
| 70-80 |
1 |
75 |
4 |
4 |
16 |
16 |
| |
79 |
|
|
-71 |
|
305 |
Here, $N = 79, A = 35, \sum\text{f}_{\text{i}}\text{u}_{\text{i}}=-71,\ \sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2=305$ and $h = 10$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)$
$\Rightarrow\overline{\text{x}}=35+10\Big(\frac{-71}{79}\Big)=26.01$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{u}_{\text{i}}\Big)^2\bigg]$
$\text{Var}(\text{X})=100\bigg[\frac{305}{79}-\Big(\frac{-71}{79}\Big)^2\bigg]=305.30$
$\therefore\text{SD}=\sqrt{\text{var}(\text{x})}=\sqrt{305.30}=17.47$ View full question & answer→Question 405 Marks
Calculate the mean deviation from the mean for the following data:
4, 7, 8, 9, 10, 12, 13, 17
Answer$\text{Mean}=\frac{1}{\text{n}}\sum|\text{x}_\text{i}|=\frac{80}{8}=10$
Calculation of mean Deviation
|
X-values
|
deviation From Mean
|
|
4
|
6
|
|
7
|
3
|
| 8 |
2 |
| 9 |
1 |
|
10
|
0
|
|
12
|
2
|
| 13 |
3 |
|
17
|
7
|
|
Total
|
24
|
We have,
$\sum|\text{x}_\text{i}-10|=\sum\text{d}_\text{i}=24$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{8}[24]=3$ View full question & answer→Question 415 Marks
A student obtained the mean and standard deviation of $100$ observations as $40$ and $5.1$ respectively. It was later found that one observation was wrongly copied as $50$, the correct figure being $40.$ Find the correct mean and $S.D.$
AnswerWe have, $\text{n} = 100,\ \overline{\text{x}}=40$ and $\sigma=5.1$
$\therefore\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_{\text{i}}$
$\Rightarrow\sum\text{x}_{\text{i}}=\text{n}\overline{\text{x}}=100\times40=4000$
$\therefore$ Incorrect $\sum\text{x}_{\text{i}}=4000$
and,
$\sigma=5.1$
$\Rightarrow\sigma^2=26.01$
$\Rightarrow\frac{1}{\text{n}}\sum\text{x}_{\text{i}}^2-(\text{Mean})^2=26.01$
$\Rightarrow\frac{1}{100}\sum\text{x}_{\text{i}}^2-1600=26.01$
$\Rightarrow\sum\text{x}_{\text{i}}^2=1626.01\times100$
$\therefore$ Incorrect $\sum\text{x}_{\text{i}}^2=162601$
When the incorrect observation 50 is replaced by 40:
We have, Incorrect $\sum\text{x}_{\text{i}}=4000$
$\therefore$ Corrected $\sum\text{x}_{\text{i}}=4000-50+40=3990$
and,
Incorrected $\sum\text{x}_{\text{i}}^2=162601$
$\therefore$ Corrected $\sum\text{x}_{\text{i}}^2=162601-50^2+40^2=161701$
Now, Corrected mean $=\frac{3990}{100}=39.90$
Corrected variance $=\frac{1}{100}\ \big(\text{Corrected}\sum\text{x}_{\text{i}}^2\big)$ - (Corrected mean)$^2$
⇒ Corrected variance $=\frac{161701}{100}-\Big(\frac{3990}{100}\Big)^2$
⇒ Corrected variance $=\frac{161701\times100-(3990)^2}{(100)^2}$
⇒ Corrected variance $=\frac{16170100-15920100}{10000}=25$
$\therefore$ Correctes standard deviation $=\sqrt{25}=5$
View full question & answer→Question 425 Marks
calculate the mean deviation about median of the following frequency distribution:
|
$x_i$
|
5
|
7
|
9
|
11
|
13
|
15
|
17
|
|
$f_i$
|
2
|
4
|
6
|
8
|
10
|
12
|
8
|
AnswerWe have to calculate mean deviation from the median. So, first we calculate the median.
|
$x_i$
|
$f_i$
|
cum. fre
|
$|d_i| = |x_i - 13|$
|
$f_i|d_i|$
|
|
5
|
2
|
2
|
8
|
16
|
|
7
|
4
|
6
|
6
|
24
|
|
9
|
6
|
12
|
4
|
24
|
|
11
|
8
|
20
|
2
|
16
|
|
13
|
10
|
30
|
0
|
0
|
|
15
|
12
|
42
|
2
|
24
|
|
17
|
8
|
50
|
4
|
32
|
|
|
N = 50
|
|
|
Total = 136
|
$\frac{\text{N}}{2}=25$
Value corresponding to 25 is Median = 13
$\text{M.D}=\frac{136}{50}=2.72$ View full question & answer→Question 435 Marks
The following are some particulars of the distribution of weights of boys and girls in a class:
| Number |
Boys
|
Girls
|
| |
100 |
50
|
| Mean weight |
60kg
|
45kg
|
|
Variance
|
9 |
4
|
Which of the distributions is more variable?
AnswerIn order to compare the variability of weight in boys and girls, we have to calculate their coefficients of variation.
Let $\sigma_1$ and $\sigma_2$ denote the standard deviations of weight in boys and girls respectively. Further,
Let $\overline{\text{X}}_1$ and $\overline{\text{X}}_2$ be the mean weight of boys and girls respectively.
We have,
$\overline{\text{X}}_1=60,\ \overline{\text{X}}_2=45$
$\sigma_1^2=9$ and $\sigma_2^2=4$
$\Rightarrow\sigma_1=\sqrt{9}=3$ and $\sigma_2=\sqrt{4}=2$
Now,
Coefficient of variation in weight in boys $=\frac{\sigma_1}{\overline{\text{x}_1}}\times100$
$=\frac{3}{60}\times100=5$
and,
Coefficient of variation in weight in girls $=\frac{\sigma_2}{\overline{\text{x}_2}}\times100$
$=\frac{2}{45}\times100=4.44$
Clearly, coefficient of variation in weight is greater in boys than in girls.
So, weights shows more variability in boys.
View full question & answer→Question 445 Marks
Find the mean deviation from the median for the following data:
|
Mark obtained
|
10
|
11
|
12
|
14
|
15
|
|
No. of students
|
2
|
3
|
8
|
3
|
4
|
Answer
|
$x_i$
|
$f_i$
|
Cum Freq
|
$|d_i| = |x_i - 12|$
|
$f_i|d_i|$
|
|
10
|
2
|
2
|
2
|
4
|
|
11
|
3
|
5
|
1
|
3
|
|
12
|
8
|
13
|
0
|
0
|
|
14
|
3
|
16
|
2
|
6
|
|
15
|
4
|
20
|
3
|
12
|
|
|
20
|
|
|
Total = 25
|
$\frac{\text{N}}{2}=10$
Median = 12
$\text{M.D}=\frac{25}{20}\approx1.25$ View full question & answer→Question 455 Marks
Find the coefficient of variation for the following data:
| Size (in cms): |
10-15
|
15-20
|
20-25
|
25-30 |
30-35 |
35-40 |
|
No. of items:
|
2
|
8
|
20
|
35 |
20 |
15 |
Answer
| $CI$ |
$f$ |
$x$ |
$\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ |
$fu$ |
$u^2$ |
$fu^2$ |
| 10-15 |
2 |
12.5 |
-2 |
-4 |
4 |
8 |
| 15-20 |
8 |
17.5 |
-1 |
-8 |
1 |
8 |
| 20-25 |
20 |
22.5 |
0 |
0 |
0 |
0 |
| 25-30 |
35 |
27.5 |
1 |
35 |
1 |
35 |
| 30-35 |
20 |
32.5 |
2 |
40 |
4 |
80 |
| 35-40 |
15 |
37.5 |
3 |
45 |
9 |
135 |
| |
100 |
|
|
108 |
|
266 |
Here, $N = 100, A = 22.5, \sum\text{f}_\text{i}\text{u}_\text{i}=108,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=266$ and $h = 5$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=22.5+5\Big(\frac{108}{100}\Big)=27.90$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=25\bigg[\frac{266}{100}-\Big(\frac{108}{100}\Big)^2\bigg]=37.34$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{37.34}=6.11$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{6.11}{27.90}\times100=21.9$ View full question & answer→Question 465 Marks
The mean and standard deviation of $20$ observation are found to be $10$ and $2$ respectively. On reacheking it was found that an observation $8$ was incorrect. Calculate the correct and standard deviation in each of the following cases:
- If wrong item is omitted.
- If it is replaced by $12.$
Answer$\text{n}=20,\overline{\text{X}}=10,\sigma=2$
$\therefore\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}$
$\Rightarrow\sum\text{x}_\text{i}=\text{n}\overline{\text{x}}=20\times10=200$
Incorrected $\sum\text{x}_\text{i}=200$
and,
$\sigma=2$
$\Rightarrow\sigma^2=4$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\text{Mean})^2=4$
$\Rightarrow\frac{1}{20}\sum{\text{x}_\text{i}}^2-100=4$
$\Rightarrow\sum{\text{x}_\text{i}}^2=104\times20$
$\Rightarrow\sum{\text{x}_\text{i}}^2=2080.$
- When $8$ is omitted from the data:
If $8$ is omitted from the data, then $19$ observation are left.
Now, Incorrected $\sum\text{x}_\text{i}=200$
$\Rightarrow$ Corrected $\sum{\text{x}_\text{i}}^2+8^2=2080$
$\Rightarrow$ Corrected $\sum{\text{x}_\text{i}}^2=2080-64$
$\Rightarrow$ Corrected $\sum\text{x}_\text{i}^2=2016$
$\therefore$ Corrected mean $=\frac{192}{19}=10.10$
$\Rightarrow$ Corrected variance $=\frac{1}{19}\big(\text{corrected}\sum{\text{x}_\text{i}}^2\big)-\big(\text{Corrected}\ \text{mean}\big)^2$
$\Rightarrow$ Corrected variance $=\frac{2016}{19}-\Big(\frac{192}{19}\Big)^2$
Corrected variance $=\frac{38304-36864}{361}=\frac{1440}{361}$
$\therefore$ Corrected standard deviation $=\sqrt{\frac{1440}{361}}=\frac{12\sqrt{10}}{19}=1.997$
- When the incorrect observation $8$ is replaced by $12:$
We have, Incorrected $\sum\text{x}_\text{i}=200$
$\therefore$ Corrected $\sum\text{x}_\text{i}=208-8+12=204$
and,
$\Rightarrow \text{Incorrected}\sum{\text{x}_\text{i}}^2=2080$
$\therefore$ Corrected $\sum{\text{x}_\text{i}}^2=2080-8^2+12^2=2160$
Now, $\therefore$ Corrected mean $=\frac{204}{20}=10.2$
Corrected variance $=\frac{1}{20}(\text{Corrected}\sum{\text{x}_\text{i}}^2)-(\text{Corrected}\ \text{mean})^2$
$\Rightarrow$ Corrected variance $=\frac{2160}{20}-\Big(\frac{204}{20}\Big)^2$
$\Rightarrow$ Corrected variance $=\frac{2160\times20-(204)^2}{(20)^2} $
$\Rightarrow$ Corrected variance $=\frac{43200-41616}{400}=\frac{1584}{400}$
$\therefore$ Corrected standard deviation $=\sqrt{\frac{1584}{400}}=\frac{\sqrt{396}}{10}=1.9899$ View full question & answer→Question 475 Marks
Find the standard deviation for the following data:
|
x
|
2 |
3
|
4
|
5
|
6
|
7 |
|
f
|
4
|
9
|
16
|
14
|
11
|
6 |
Answer
|
$x_i$
|
$f_i$
|
$f_ix_i^2$
|
|
2
|
4
|
16
|
|
3
|
9
|
81
|
|
4
|
16
|
256
|
|
5
|
14
|
350
|
|
6
|
11
|
396
|
|
7
|
6
|
294
|
| |
N = 60 |
Total = 1393 |
Mean $=\frac{8+27+64+70+66+42}{60}=\frac{277}{60}=4.62$
Var $=\frac{1393}{60}-(4.62)^2=1.88$
SD $=\sqrt{1.88}=1.37$ View full question & answer→Question 485 Marks
Find the numberof observation lying between $\overline{\text{X}}-\text{M.D. }$and $\overline{\text{X}}-\text{M.D. }$ is the mean deviation from the mean.
$34, 66, 30, 38, 44, 50, 40, 60, 42, 51$
AnswerLet $\overline{\text{x}}$ be the mean of the data set.
$\overline{\text{x}}=\frac{34+66+30+38+44+50+40+60+42+51}{10}=45.5$
$\text{MD}=\frac{1}{\text{n}}\sum_\limits{\text{i}=1}^\text{n}|\text{d}_\text{i}|,\text{where}|\text{d}_\text{i}|=|\text{x}_\text{i}-\overline{\text{x}}|$
|
$x_i$
|
$|d_i| = |x_i - 45.5|$
|
|
34
|
11.5
|
|
66
|
20.5
|
|
30
|
15.5
|
|
38
|
7.5
|
|
44
|
1.5
|
|
50
|
4.5
|
|
40
|
5.5
|
|
60
|
14.5
|
|
42
|
3.5
|
|
51
|
5.5
|
|
Total
|
90
|
$\text{MD}=\frac{1}{10}\times90=9$
$\overline{\text{x}}$ - M.D. = 45.5 - 9 = 36.5
Also, $\overline{\text{x}}$ + M.D. = 45.5 + 9 = 54.5
Hence, there are 6 observations between 36.5 and 54.5. View full question & answer→Question 495 Marks
While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
AnswerMean = 45
Variance = 16
n = 10
$\sum\text{x}_{\text{i}}=450$
Corrected Sum = 450 - 52 + 25 = 423
Corrected Mean = 42.3
Variance = 16
$16=\frac{\sum\text{x}_{\text{i}}^2}{10}-(45)^2$
Incorrected $\sum\text{x}_{\text{i}}^2=20410$
Corrected $\sum\text{x}_{\text{i}}^2=$
Incorrect $\sum\text{x}_{\text{i}}^2$ - (Sum of squares of incorrect value) + (Sum of squares of corrected value)
Corrected $\sum\text{x}_{\text{i}}^2=20410-2704+625=18331$
Corrected $\sigma=\sqrt{\frac{\text{Corrected}\sum\text{x}_\text{i}^2}{\text{n}}-(\text{Corrected Mean})^2}$
Corrected $\sigma=\sqrt{\frac{18331}{10}-(42.3)^2}=6.62$
Corrected Variance = 6.62 × 6.62 = 43.82
View full question & answer→Question 505 Marks
The mean and standard deviation of $6$ observation are $8$ and $4$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observation.
AnswerMean = $\overline{\text{X}}=8$
$n = 6$
$\sigma=\text{S.D}=4$
If $x_1, x_2, ......x_6 $are the given observation
$\overline{\text{X}}=\frac{1}{\text{n}}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
$\Rightarrow8=\frac{1}{6}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
Let $u_1, u_2.....u_6$ be the new observation
$\Rightarrow u_i = 3x_i $(for $i = 1, 2, 3...6)$
⇒ Mean of new observation $=\overline{\text{U}}=\frac{1}{\text{n}}\times\sum_\limits{\text{i}=1}^6\text{u}_\text{i}$
$=\frac{1}{6}\times\sum_\limits{\text{i}=1}^63\text{x}_\text{i}$
$=3\times\frac{1}{6}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
$=3\ \overline{\text{X}}$
$=3\times8$
$=24$
$\text{SD}=\sigma_\text{x}=4$
${\sigma_\text{x}}^2$ = Variance X
$\therefore$ Variance X = 16
$\Rightarrow\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{x}_\text{i}-\overline{\text{X}})^2=16.....(1)$
$\text{Variance}\ (\text{U})={\sigma_\text{u}}^2=\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{u}_\text{i}-\overline{\text{U}})^2$
$=\frac{1}{6}\times\sum_\limits{\text{i}=1}^6(3\text{x}_\text{i}-3\overline{\text{X}})^2$
$=3^2\times\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{x}_\text{i}-\overline{\text{X}})^2$
$=9\times16$
$\sigma_\text{u}=\sqrt{\text{Variance}\ (\text{U})}$
$=\sqrt{9\times16}$
$=12$
View full question & answer→Question 515 Marks
Find the numberof observation lying between $\overline{\text{X}}-\text{M.D. }$and $\overline{\text{X}}+\text{M.D. }$ is the mean deviation from the mean.
$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$
AnswerLet $\overline{\text{x}}$ be the mean of the data set.
$\overline{\text{x}}=\frac{22+24+30+27+29+31+25+28+41+42}{10}=29.9$=
|
$x_i$
|
$|d_i| = |x_i - 29.9|$
|
|
22
|
7.9
|
|
24
|
5.9
|
|
30
|
0.1
|
|
27
|
2.9
|
|
29
|
0.9
|
|
31
|
1.1
|
|
25
|
4.9
|
|
28
|
1.9
|
|
41
|
11.9
|
|
42
|
12.1
|
|
Total
|
48.8
|
$\text{MD}=\frac{1}{10}\times48.8=4.88$
$\overline{\text{x}}$ - M.D. = 29.9 - 4.88 = 25.02,
and, $\overline{\text{x}}$ + M.D. = 29.9 + 4.88 = 34.78
There are 5 observation between 25.02 and 34.78. View full question & answer→Question 525 Marks
Show that the two formula for the standard deviation of ungrouped data
$\sigma=\sqrt{\frac{1}{\text{n}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}}\text{and}\ \sigma^{'}=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\overline{\text{X}^2}}$ are equivalent, where $ \overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
Answer$\sigma=\sqrt{\frac{1}{\text{n}}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}$
$=\sqrt{\frac{1}{\text{n}}\sum\big({\text{x}_\text{i}^2}-2{\text{x}_\text{i}}\overline{\text{X}}+\overline{\text{X}}\big)^2}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}^2_\text{i}}-\frac{1}{\text{n}}\sum2\text{x}_\text{i}\overline{\text{X}}+\frac{1}{\text{n}}\sum\overline{{\text{X}}}^2}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}}^2_\text{i}-\frac{1}{\text{n}}\times2\overline{\text{X}}\sum\text{x}_\text{i}+\frac{1}{\text{n}}\times\overline{\text{X}}^2\sum1}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}^2_\text{i}}-\frac{1}{\text{n}}\times2\overline{\text{X}}\times\text{n}\overline{\text{X}}+\frac{1}{\text{n}}\times\overline{\text{X}}^2\times\text{n}}$
$=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-2\overline{\text{X}}^2+\overline{\text{X}}^2}$
$=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-\overline{\text{X}}^2}$
$=\sigma^{'}$
Hence, the formula $\sigma=\sqrt{\frac{1}{\text{n}}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}\ \text{and}\ \sigma^{'}=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-\overline{\text{X}}^2}$ are equivalent, where $\overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
View full question & answer→Question 535 Marks
The mean and variance of $8$ observation are $9$ and $9.25$ respectively. If six of the observation are $6, 7, 10, 12, 12$ and $13$, find the remaining two observation.
AnswerLet x and y be the remaining two observations. Then,
Mean = 9
$\Rightarrow\frac{6+7+10+12+12+13+\text{x}+\text{y}}{8}=9$
$\Rightarrow 60 + x + y =72$
$\Rightarrow x + y = 12.........(i)$
Variance $= 9.25$
$\Rightarrow\frac{1}{8}\big(6^2+7^2+10^2+12^2+12^2+13^2+\text{x}^2+\text{y}^2\big)-(\text{Mean})^2=9.25$
$\Rightarrow\frac{1}{8}\big(36+49+100+144+144+169+\text{x}^2+\text{y}^2\big)-81=9.25$
$\Rightarrow 642 + x^2 + y^2 = 722$
$\Rightarrow x^2 + y^2 = 80........(ii)$
Now, $(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$
$\Rightarrow 144 + (x - y)^2= 2 \times 80$
$\Rightarrow x - y = 16$
$\Rightarrow\text{x}-\text{y}=\pm4$
if x - y = 4,
then $x + y =12$ and $x - y = 4$
$\Rightarrow x = 8, y = 4$
if $x - y = -4,$
then $x + y =12$ and $x - y = -4$
$\Rightarrow x = 4, y = 8$
Hence, the remaining two observation are 4 and 8.
View full question & answer→Question 545 Marks
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observation were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observation were omitted.
AnswerWe have, $\text{n}=100,\overline{\text{x}}=20\ \text{and}\ \sigma=3$
Since $\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}$
$\Rightarrow\sum\text{x}_\text{i}=\text{n}\overline{\text{x}}=20\times100=2000$
$\Rightarrow\text{Incorrect}\ \sum\text{x}_\text{i}=2000$
and,
$\sigma=3$
$\Rightarrow\sigma^2=9$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\text{Mean})^2=9$
$\Rightarrow\frac{1}{100}\sum{\text{x}_\text{i}}^2-400=9$
$\Rightarrow\sum{\text{x}_\text{i}}^2=409\times100$
$\Rightarrow\text{Incorrect}\sum{\text{x}_\text{i}}^2=40900.$
When the incorrect observation 21, 21, 18 are omitted from the data:
n = 97
Now, $\text{Incorrect}\sum\text{x}_\text{i}=2000$
$\Rightarrow\sum{\text{x}_\text{i}}=2000-21-21-18=1940$
and,
$\text{Incorrect}\sum{\text{x}_\text{i}}^2=40900$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-21^2-21^2-18^2$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-1206$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=39694$
$\therefore\text{Corrected}\ \text{mean}=\frac{1940}{97}=20$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{1}{97}\big(\text{Corrected}\sum{\text{x}_\text{i}}^2\big)-\big(\text{Corrected}\ \text{mean}\big)^2$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{39694}{97}-(20)^2=409.22-400=9.22$
$\therefore\text{Corrected}\ \text{standard}\ \text{deviation}=\sqrt{9.22}=3.04$
View full question & answer→Question 555 Marks
Calculate the mean, variance and standard deviation of the following frequency distribution.
| Class: |
1-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
| Frequency: |
11 |
29 |
18 |
4 |
5 |
3 |
Answer
|
Class interval
|
$f_i$
|
Mid-value $x_i$
|
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-35}{10}$
|
$f_iu_i$
|
$u_i^2$
|
$f_iu_i^2$
|
|
0-10
|
11
|
5
|
-3
|
-33
|
9
|
99
|
|
10-20
|
29
|
15
|
-2
|
-58
|
4
|
116
|
|
20-30
|
18
|
25
|
-1
|
-18
|
1
|
18
|
|
30-40
|
4
|
35
|
0
|
0
|
0
|
0
|
|
40-50
|
5
|
45
|
1
|
5
|
1
|
5
|
|
50-60
|
3
|
55
|
2
|
6
|
4
|
12
|
|
|
$\text{N}=\sum\text{f}_\text{i}=70$
|
|
|
$\sum\text{f}_\text{i}\text{u}_\text{i}=-98$
|
|
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=250$
|
$N = 70, \sum\text{f}_\text{i}\text{u}_\text{i}^2=250, A = 35$ and $h = 10$
$\text{Mean}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\text{Mean}=35+10\Big(\frac{-98}{70}\Big)=-21$
$\text{Var}\big(\text{X}\big)=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\bigg\{\Big(\frac{1}{70}\times250\Big)-\Big(\frac{1}{70}\times(-98)\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\big\{3.57-1.96\big\}=161$
$\text{SD}=\sqrt{\text{Var}(\text{X})}=\sqrt{161}=12.7$ View full question & answer→Question 565 Marks
calculate the mean deviation from the mean for the following data:
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer$\text{Mean}=\frac{1}{\text{n}}\sum|\text{x}_\text{i}|=\frac{168}{12}=14$
Calculation of Mean Deviation
|
X-values
|
Deviation From Mean
|
|
13
|
1
|
|
17
|
3
|
|
16
|
2
|
|
14
|
0
|
|
11
|
3
|
|
13
|
1
|
|
10
|
4
|
|
16
|
2
|
|
11
|
3
|
|
18
|
4
|
|
12
|
2
|
|
17
|
3
|
|
Total
|
28
|
We have,
$\sum|\text{x}_\text{i}-14|=\sum\text{d}_\text{i}=28$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{12}[28]=2.33$ View full question & answer→Question 575 Marks
Table below shows the frequency $f$ with which $'x'$ alpha particles radiated from a diskette:
| x |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| f |
51 |
203 |
383 |
525 |
532 |
408 |
273 |
139 |
43 |
27 |
10 |
4 |
2 |
Calculate the mean and variance. AnswerMean, $\overline{\text{x}}=\frac{\sum\text{f}_{\text{i}}\text{x}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{10078}{2600}=3.88$
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
$\text{x}_{\text{i}}-\overline{\text{X}}$
|
$\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
|
$\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
|
|
0
|
51
|
0
|
-3.88
|
15.05
|
767.55
|
|
1
|
203
|
203
|
-2.88
|
8.29
|
1682.87
|
|
2
|
383
|
766
|
-1.88
|
3.53
|
1351.99
|
|
3
|
525
|
1575
|
-0.88
|
0.77
|
404.25
|
|
4
|
532
|
2128
|
0.12
|
0.014
|
7.448
|
|
5
|
408
|
2040
|
1.12
|
1.25
|
510
|
|
6
|
273
|
1638
|
2.12
|
4.49
|
1225.77
|
|
7
|
139
|
973
|
3.12
|
9.73
|
1352.47
|
|
8
|
43
|
344
|
4.12
|
16.97
|
729.71
|
|
9
|
27
|
243
|
5.12
|
26.21
|
707.67
|
|
10
|
10
|
100
|
6.12
|
37.45
|
374.5
|
|
11
|
4
|
44
|
7.12
|
50.69
|
202.76
|
|
12
|
2
|
24
|
8.12
|
65.93
|
131.86
|
|
|
$\sum\text{f}_{\text{i}}=\text{N}=2600$
|
$\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=10078$
|
|
|
$\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2=9448.848$
|
Variance, $\sigma^2=\frac{\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2}{{\text{N}}}=\frac{9448.848}{2600}=3.63$ View full question & answer→Question 585 Marks
Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
AnswerMean = 40
SD = 10
n = 100
$\sum\text{x}_{\text{i}}=40\times100=4000$
Corrected Sum = 4000 - 30 - 70 + 3 + 27 = 3930
Corrected Mean = $\frac{3930}{100}=39.3$
Variance = 100
$100=\frac{\sum\text{x}_{\text{i}}^2}{100}-(40)^2$
Incorrected $\sum\text{x}_{\text{i}}^2=170000$
Corrected $\sum\text{x}_{\text{i}}^2=$
Incorrect $\sum\text{x}_{\text{i}}^2$ - (Sum of squares of incorrect value) + (Sum of squares of corrected value)
Corrected $\sum\text{x}_{\text{i}}^2=170000-(900+4900)+(9+729)$
Corrected $\sum\text{x}_{\text{i}}^2=164938$
Corrected $\sigma=\sqrt{\frac{\text{Corrected}\sum\text{x}_\text{i}^2}{\text{n}}-(\text{Corrected Mean})^2}$
Corrected $\sigma=\sqrt{\frac{164938}{100}-(39.3)^2}=10.24$
View full question & answer→Question 595 Marks
Find the mean, and standard deviation for the following data:
|
Marks:
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
13
|
14
|
15
|
16
|
|
Frequency:
|
1
|
6
|
6
|
8
|
8
|
2
|
2
|
3
|
0
|
2
|
1
|
0
|
0
|
0
|
1
|
Answer
| $x_i$ |
$f_i$ |
$f_ix_i$ |
$f_ix_i{}^2$ |
| 2 |
1 |
2 |
4 |
| 3 |
6 |
18 |
54 |
| 4 |
6 |
24 |
96 |
| 5 |
8 |
40 |
200 |
| 6 |
8 |
48 |
288 |
| 7 |
2 |
14 |
98 |
| 8 |
2 |
16 |
128 |
| 9 |
3 |
27 |
243 |
| 10 |
0 |
0 |
0 |
| 11 |
2 |
22 |
242 |
| 12 |
1 |
12 |
144 |
| 13 |
0 |
0 |
0 |
| 14 |
0 |
0 |
0 |
| 15 |
0 |
0 |
0 |
| 16 |
1 |
16 |
256 |
| |
N = 40 |
Total = 239 |
Total = 1753 |
Mean $=\frac{239}{40}=5.975$
Var $=\frac{1753}{40}-(5.975)^2=8.12$
SD $=\sqrt{8.12}=2.85$ View full question & answer→Question 605 Marks
Following are the marks obtained, out of $100$, by two students Ravi and Hashina in $10$ tests:
| Ravi: |
25 |
50 |
45 |
30 |
70 |
42 |
36 |
48 |
35 |
60 |
| Hashina: |
10 |
70 |
50 |
20 |
95 |
55 |
42 |
60 |
48 |
80 |
Who is more intelligent and who is more consistent? AnswerFor Ravi:
|
Marks $(x_i)$
|
$d_i = x_i - 45$
|
$d_i{}^2$
|
|
25
|
-20
|
400
|
|
50
|
5
|
25
|
|
45
|
0
|
0
|
|
30
|
-15
|
225
|
|
70
|
25
|
625
|
|
42
|
-3
|
9
|
|
36
|
-9
|
81
|
|
48
|
3
|
9
|
|
35
|
-10
|
100
|
|
60
|
15
|
225
|
|
|
$\sum\text{d}_\text{i}=-9$
|
$\sum\text{d}_\text{i}^2=1699$
|
Mean, $\overline{\text{X}}_\text{R}=\text{A}+\frac{\sum\text{d}_\text{i}}{10}=45+\frac{(-9)}{10}=44.1$
Standard deviation, $\sigma_\text{R}$
$=\sqrt{\frac{\sum\text{d}_\text{i}^2}{10}-\Big(\frac{\sum\text{d}_\text{i}}
{10}\Big)^2}$
$=\sqrt{\frac{1699}{10}-\Big(\frac{-9}{10}\Big)^2}$
$=\sqrt{169.09}=13.003$
Coefficient of variation $=\frac{\sigma_\text{R}}{\overline{\text{X}}_\text{R}}\times100$
$=\frac{13.003}{44.1}\times100$
$=29.49$
For Hashina:
|
Marks $(x_i)$
|
$d_i = x_i - 55$
|
$d_i{}^2$
|
|
10
|
-45
|
2025
|
|
70
|
15
|
625
|
|
50
|
-5
|
25
|
|
20
|
-35
|
1225
|
|
95
|
40
|
1600
|
|
55
|
0
|
0
|
|
42
|
-13
|
169
|
|
60
|
5
|
25
|
|
48
|
-7
|
49
|
|
80
|
25
|
625
|
|
|
$\sum\text{d}_\text{i}=-20$
|
$\sum\text{d}_\text{i}^2=6368$
|
Mean, $\overline{\text{X}}_\text{H}=\text{A}+\frac{\sum\text{d}_\text{i}}{10}=55+\frac{(-20)}{10}=53$
Standard deviation, $\sigma_\text{H}$
$=\sqrt{\frac{\sum\text{d}_\text{i}^2}{10}-\Big(\frac{\sum\text{d}_\text{i}}{10}\Big)^2}$
$=\sqrt{\frac{6368}{10}-\Big(\frac{-20}{10}\Big)^2}$
$=\sqrt{632.8}=25.16$
Coefficient of variation $=\frac{\sigma_\text{H}}{\overline{\text{X}}_\text{H}}\times100=\frac{25.16}{53}\times100=47.47$
Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent. View full question & answer→Question 615 Marks
Find the mean, and standard deviation for the following data:
| Year render: |
10 |
20 |
30 |
40 |
50 |
60 |
| No. of persons(cumulative): |
15 |
32 |
51 |
78 |
97 |
109 |
Answer
| $x$ |
Cum Freq |
$f_i$ |
$f_ix_i$ |
$f_ix_i^2$ |
| 10 |
15 |
15 |
150 |
1500 |
| 20 |
32 |
17 |
340 |
6800 |
| 30 |
51 |
19 |
570 |
17100 |
| 40 |
78 |
27 |
1080 |
43200 |
| 50 |
97 |
19 |
950 |
47500 |
| 60 |
109 |
12 |
720 |
43200 |
| |
|
N = 109 |
Total = 3810 |
Total = 159300 |
Mean $=\frac{3810}{109}=34.95$
Var $=\frac{159300}{109}-(34.95)^2=239.96$
SD $=\sqrt{239.96}=15.49$ View full question & answer→Question 625 Marks
From the prices of shares $X$ and $Y$ given below: find out which is more stable in value:
| x |
35 |
54 |
52 |
53 |
56 |
58 |
52 |
50 |
51 |
49 |
| y |
108 |
107 |
105 |
105 |
106 |
107 |
104 |
103 |
104 |
101 |
Answer
| $x$ |
$d = (x - Mean)$ |
$d^2$ |
| 35 |
-13 |
169 |
| 24 |
-24 |
576 |
| 52 |
4 |
16 |
| 53 |
5 |
25 |
| 56 |
8 |
64 |
| 58 |
10 |
100 |
| 52 |
4 |
16 |
| 50 |
2 |
4 |
| 51 |
3 |
9 |
| 49 |
1 |
1 |
| 480 |
|
980 |
$\overline{\text{x}}=\frac{1}{\text{N}}\sum\text{x}_\text{i}=\frac{1}{10}[480]=48$
$\text{Var}(\text{X})=\frac{1}{\text{N}}\Big\{\sum\big(\text{x}_\text{i}-\overline{\text{x}}\big)^2\Big\}=\frac{1}{10}(980)=98$
$\therefore\text{S.D.}(\text{X})=\sqrt{\text{Var}(\text{X})}=\sqrt{98}=9.9$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}_1}\times100=\frac{9.9}{48}\times100=20.6$
| $x$ |
$d = (x - Mean)$ |
$d^2$ |
| 35 |
-13 |
169 |
| 24 |
-24 |
576 |
| 52 |
4 |
16 |
| 53 |
5 |
25 |
| 56 |
8 |
64 |
| 58 |
10 |
100 |
| 52 |
4 |
16 |
| 50 |
2 |
4 |
| 51 |
3 |
9 |
| 49 |
1 |
1 |
| 480 |
|
980 |
$\overline{\text{x}}=\frac{1}{\text{N}}\sum\text{x}_\text{i}=\frac{1}{10}[1050]=105$
$\text{Var}(\text{X})=\frac{1}{\text{N}}\Big\{\sum\big(\text{x}_\text{i}-\overline{\text{x}}\big)^2\Big\}=\frac{1}{10}(40)=4$
$\therefore\text{S.D}(\text{X})=\sqrt{\text{Var}(\text{X})}=\sqrt{4}=2$
Coefficient of variation for shares $\text{Y}=\frac{\text{S.D.}}{\overline{\text{x}}_1}\times100=\frac{2}{105}\times100=1.90$
Since the coefficient of variation for share Y is smaller than the coefficient of variation for share X, they are more stable. View full question & answer→Question 635 Marks
From the data given below state which group is more variable, $G_1$ or $G_2$?
| Marks |
10-20
|
20-30
|
30-40
|
40-50 |
50-60 |
60-70 |
70-80 |
|
Group $G_1$
|
9
|
17
|
32
|
33 |
40 |
10 |
9 |
|
Group $G_2$
|
10
|
20
|
30
|
25 |
43 |
15 |
7 |
AnswerLet's first find the coefficient of variable for Group $G_1$
| $CI$ |
$f$ |
$x$ |
$\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ |
$fu$ |
$u^2$ |
$fu^2$ |
| 10-20 |
9 |
15 |
-3 |
-27 |
9 |
81 |
| 20-30 |
17 |
25 |
-2 |
-34 |
4 |
68 |
| 30-40 |
32 |
35 |
-1 |
-32 |
1 |
32 |
| 40-50 |
33 |
45 |
0 |
0 |
0 |
0 |
| 50-60 |
40 |
55 |
1 |
40 |
1 |
40 |
| 60-70 |
10 |
65 |
2 |
20 |
4 |
40 |
| 70-80 |
9 |
75 |
3 |
27 |
9 |
81 |
| |
150 |
|
|
-6 |
|
342 |
Here, $N = 150, A = 45, \sum\text{f}_\text{i}\text{u}_\text{i}=-6,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=342$ and $h = 10$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=45+10\Big(\frac{-6}{150}\Big)=44.6$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=100\bigg[\frac{342}{150}-\Big(\frac{-6}{150}\Big)^2\bigg]=227.84$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{227.84}=15.09$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{15.09}{44.6}\times100=33.83$
Now, Let's first find the coefficient of variable for Group $G_2$
| $CI$ |
$f$ |
$x$ |
$\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ |
$fu$ |
$u^2$ |
$fu^2$ |
| 10-20 |
10 |
15 |
-3 |
-30 |
9 |
902 |
| 20-30 |
20 |
25 |
-2 |
-40 |
4 |
80 |
| 30-40 |
30 |
35 |
-1 |
-30 |
1 |
30 |
| 40-50 |
25 |
45 |
0 |
0 |
0 |
0 |
| 50-60 |
43 |
55 |
1 |
43 |
1 |
43 |
| 60-70 |
15 |
65 |
2 |
30 |
4 |
60 |
| 70-80 |
7 |
75 |
3 |
21 |
9 |
63 |
| |
150 |
|
|
-6 |
|
366 |
Here, $N = 150, A = 45, \sum\text{f}_\text{i}\text{u}_\text{i}=-6,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=366$ and $h = 10$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=45+10\Big(\frac{-6}{150}\Big)=44.6$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=100\bigg[\frac{366}{150}-\Big(\frac{-6}{150}\Big)^2\bigg]=243.84$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{227.84}=15.62$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{15.09}{44.6}\times100=35.02$
$\therefore$ Group $G_2$ is more variable. View full question & answer→Question 645 Marks
The lengths $($in $cm)$ of $10$ rods in a shop are given below:
$40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2$
- Find mean deviation from median
- Find mean deviation from the mean also.
AnswerFirst arrange the given numbers in assending order
write these number in assending order
$40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2$
we get $15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0$
Clearly, $\text{Median}=\frac{40.0+52.3}{2}=46.15$
Let $\overline{\text{x}}$ be the mean of given data , we get
$\overline{\text{x}}=\frac{15.2+27.9+ 30.2+32.5+40.0+52.3+52.8+55.2+72.9+79.0}{10}=45.8$
Calculation of mean Deviations from mean and median
| $x_i$ |
$|d_i| = |x_i - 46.15|$ |
$|d_i| = |x_i - 45.8|$ |
| $40.0$ |
$6.15$ |
$5.8$ |
| $52.3$ |
$6.15$ |
$6.5$ |
| $55.2$ |
$9.05$ |
$9.4$ |
| $72.9$ |
$26.75$ |
$27.1$ |
| $52.8$ |
$6.65$ |
$7$ |
| $79.0$ |
$32.85$ |
$33.2$ |
| $32.5$ |
$13.65$ |
$13.3$ |
| $15.2$ |
$30.95$ |
$30.6$ |
| $27.9$ |
$19.25$ |
$17.9$ |
| $30.2$ |
$15.95$ |
$15.6$ |
| Total |
$167.4$ |
$166.4$ |
- $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{167.4}{10}=16.74$
- $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{166.4}{10}=16.64$
View full question & answer→Question 655 Marks
Life of bulbs produced by two factories $A$ and $B$ are given below:
| Length of life (in hours): |
550-650 |
650-750 |
750-850 |
850-950 |
950-1050 |
| Factory A: (Number of bulbs) |
10 |
22 |
52 |
20 |
16 |
| Factory B: (Number of bulbs) |
8 |
60 |
24 |
16 |
12 |
The bulbs of which factory are more consistent from the point of view of length of life? AnswerFactor A:
|
Length of life
|
Mid value $x_i$
|
$f_i$
|
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-800}{100}$
|
$f_iu_i$
|
$f_iu_i{}^2$
|
|
550-650
|
600
|
10
|
-2
|
-20
|
40
|
|
650-750
|
700
|
22
|
-1
|
-22
|
22
|
|
750-850
|
800
|
52
|
0
|
0
|
0
|
|
850-950
|
900
|
20
|
1
|
20
|
20
|
|
950-1050
|
1000
|
16
|
2
|
32
|
64
|
|
|
|
$\text{N}=\sum\text{f}_\text{i}=120$
|
|
$\sum\text{f}_\text{i}\text{u}_\text{i}=10$
|
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=146$
|
$N = 120, \sum\text{f}_\text{i}\text{u}_\text{i}=10,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=146,$ A = 800 and $h = 100$
$\overline{\text{x}}_\text{A}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)=800+100\Big(\frac{10}{120}\Big)=808.33$
$\sigma_\text{A}^2=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000\bigg\{\Big(\frac{1}{120}\times146\Big)-\Big(\frac{1}{120}\times(10)\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000(1.2166-0.0069)=12097$
$\Rightarrow\sigma_\text{A}^2=\sqrt{12097}=109.98\approx110$
Factor B:
|
Length of life
|
Mid value $x_i$
|
$f_i$
|
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-800}{100}$
|
$f_iu_i$
|
$f_iu_i^2$
|
|
550-650
|
600
|
8
|
-2
|
-16
|
32
|
|
650-750
|
700
|
60
|
-1
|
-60
|
60
|
|
750-850
|
800
|
24
|
0
|
0
|
0
|
|
850-950
|
900
|
16
|
1
|
16
|
16
|
|
950-1050
|
1000
|
12
|
2
|
12
|
48
|
|
|
|
$\text{N}=\sum\text{f}_\text{i}=120$
|
|
$\sum\text{f}_\text{i}\text{u}_\text{i}=-48$
|
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=156$
|
$N = 120, \sum\text{f}_\text{i}\text{u}_\text{i}=-48,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=156,$ A = 800 and $h = 100$
$\overline{\text{x}}_\text{B}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)=800+100\Big(\frac{-48}{120}\Big)=760$
$\sigma_\text{B}^2=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000\bigg\{\Big(\frac{1}{120}\times156\Big)-\Big(\frac{1}{120}\times(-48)\Big)^2\bigg\}$
$\sigma_\text{B}^2=10000(1.3-0.16)=11400$
$\Rightarrow\sigma_\text{B}=\sqrt{11400}=106.77\approx107$
Bulbs of factory A are more consistent from the point of view of life. View full question & answer→Question 665 Marks
Find the standard deviation for the following distribution:
| x |
4.5 |
14.5 |
24.5 |
34.5 |
44.5 |
54.5 |
64.5 |
| f |
1 |
5 |
12 |
22 |
17 |
9 |
4 |
Answer
| x |
f |
fx |
x-mean |
(x-mean)$^2$ |
f(x-mean)$^2$ |
| 4.5 |
1 |
4.5 |
-33.14 |
1098.45 |
1098.45 |
| 14.5 |
5 |
72.5 |
-23.14 |
535.59 |
2677.96 |
| 24.5 |
12 |
294 |
-13.14 |
172.73 |
2072.82 |
| 34.5 |
22 |
759 |
-3.14 |
9.88 |
217.31 |
| 44.5 |
17 |
756.5 |
6.86 |
47.02 |
799.35 |
| 54.5 |
9 |
490.5 |
16.86 |
284.16 |
2557.47 |
| 64.5 |
4 |
258 |
26.86 |
721.31 |
2885.22 |
| |
N = 70 |
2635 |
|
|
12308.57 |
Here, N = 70, $\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=2635$
$\therefore\overline{\text{x}}=\frac{1}{\text{N}}\big(\sum\text{f}_{\text{i}}\text{x}_{\text{i}}\big)=\frac{2635}{70}=37.64$
we have, $\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2=12308.57$
$\therefore\text{ver}(\text{x})=\frac{1}{\text{N}}\Big[\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2\Big]\\=\frac{12308.57}{70}=175.84$
$\text{S.D.}=\sqrt{\text{ver}(\text{x})}=\sqrt{175.84}=13.26$ View full question & answer→Question 675 Marks
The mean of 5 observation is 4.4 and their variance is 8.24. If three of the observation are 1, 2 and 6, find the other two observation.
AnswerLet the other two be x and y
1 + 2 + 6+ x + y = 5* 4.4 because of the mean
x + y = 13
Variance $=\frac{[(1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(\text{x}-4.4)^2+(\text{y}-4.4)^2]}{5}$
Hence
11.56 + 5.76 + 2.56 + (x - 4.4)^2 + (y4.4)^2 = 41.2
(x - 4.4)^2 + (y - 4.4)^2 = 21.32
Solve simuitaneously
(x - 4.4)^2 + (13 - x - 4.4)^2 = 21.32
(x- 4.4)^2 + (8.6 - x)^2 = 21.32
x^2 - 8.8x + 19.36 + 73.96 - 17.2x + x^2 = 21.32
2x^2 - 26x + 72 = 0
x^2 - 13x + 36 = 0
(x - 4)(x - 9) = 0
x = 4 or x = 9
If x = 4, y = 9 and
The other two observation are4 and 9.
View full question & answer→Question 685 Marks
Find the mean and variance of frequency distribution given below:
| $x_i$ |
$1\leq\text{x}<3$ |
$3\leq\text{x}<5$ |
$5\leq\text{x}<7$ |
$7\leq\text{x}<10$ |
| $f_i$ |
6 |
4 |
5 |
1 |
Answer
|
$x_i$
|
Midpoint value $(y_i)$
|
$y_i^2$
|
$f_i$
|
$f_iy_i$
|
$f_iy_i^2$
|
|
1-3
|
2
|
4
|
6
|
12
|
24
|
|
3-5
|
4
|
16
|
4
|
16
|
64
|
|
5-7
|
6
|
36
|
5
|
30
|
180
|
|
7-10
|
8.5
|
72.25
|
1
|
8.5
|
72.25
|
|
|
|
|
$\text{N}=\sum\text{f}_\text{i}=16$
|
$\sum\text{f}_\text{i}\text{y}_\text{i}=66.5$
|
$\sum\text{f}_\text{i}\text{y}_\text{i}^2=340.25$
|
Therfore,
$\text{Mean}=\frac{\sum\text{f}_{\text{i}}\text{y}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{66.5}{16}=4.16$
$\text{Variance}=\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}\Big)^2$
$\text{Variance}=\frac{1}{16}\times340.25-\Big(\frac{1}{16}\times66.5\Big)^2$
$\text{Variance}=21.26-17.22=4.04$ View full question & answer→Question 695 Marks
The mean and standard deviation of $100$ observation were calculated as $40$ and $5.1$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?
AnswerWe have,
$\text{n}=100,\overline{\text{X}}=40,\sigma=5.1$
$\therefore\overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}=\overline{\text{X}}=100\times40=4000.$
Corrected $\sum\text{x}_\text{i}$ = Incorrected $\sum\text{x}_\text{i}$ - (sum of incorrect values) + (sum of correct values)
= 4000 -50 + 40 = 3990
$\therefore\text{Corrected}\ \text{mean}=\frac{\text{corrected}\sum\text{x}_\text{i}}{\text{n}}=\frac{3990}{100}=39.9$
Now $\sigma=5.1$
$\Rightarrow5.1^2=\frac{1}{100}\Big(\sum{\text{x}_\text{i}}^2\Big)-\Big(\frac{1}{100}\sum\text{x}_\text{i}\Big)^2$
$\Rightarrow26.01=\frac{1}{100}\Big(\sum{\text{x}_\text{i}}^2\Big)-\Big(\frac{4000}{100}\Big)^2$
$\Rightarrow26.01=\frac{1}{100}\Big(\sum{\text{x}_\text{i}}^2\Big)-1600$
$\sum{\text{x}_\text{i}}^2=100\times1626.01=162601$
Incorrect $\sum{\text{x}_\text{i}}^2=162601$
corrected $\sum{\text{x}_\text{i}}^2$ = $ \big(\text{incorrected}\sum{\text{x}_\text{i}}^2\big)$ - (sum of squers of incorrect values) + (sum of squers of correct values)
$= 162601 - (50)^2+ (40)^2= 161701$
so, Corrected $\sigma=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\Big(\frac{1}{\text{n}}\sum\text{x}_\text{i}}\Big)^2=\sqrt{\frac{161701}{100}-\Big(\frac{3990}{100}}\Big)^2$
$=\sqrt{1617.01-1592.01}=5$
View full question & answer→Question 705 Marks
The weight of coffee in $70$ jars is shown in the following table:
| Weight (in grams): |
$200-201 |
$201-202 |
$202-203 |
$203-204 |
$204-205 |
$205-206 |
| Frequency: |
$13$ |
$27$ |
$18$ |
$10$ |
$1$ |
$1$ |
Determine the variance and standard deviation of the above distribution. Answer
|
Weight (in grams)
|
Mid-values $(x_i)$
|
Frequency$(f_i)$
|
$d_i = x_i - 202.5$
|
$d_i^2$
|
$f_id_i$
|
$f_id_i^2$
|
|
$200-201$
|
$200.5$
|
$13$
|
$-2$
|
$4$
|
$-26$
|
$52$$
|
|
$201-202
|
$201.5$
|
$27$
|
$-1$
|
$1$
|
$-27$
|
$27$
|
|
$202-203
|
$202.5$
|
$18$
|
$0$
|
$0$
|
$0$
|
$0$
|
|
$203-204$
|
$203.5$
|
$10$
|
$1$
|
$1$
|
$10$
|
$10$
|
|
$204-205$
|
$204.5$
|
$1$
|
$2$
|
$4$
|
$2$
|
$4$
|
|
$205-206$
|
$205.5$
|
$1$
|
$3$
|
$9$
|
$3$
|
$9$
|
|
|
|
$\text{N}=\sum\text{f}_\text{i}=70$
|
|
|
$\sum\text{f}_\text{i}\text{d}_\text{i}=-38$
|
$\sum\text{f}_\text{i}\text{d}_\text{i}^2=102$
|
Now,
Variance, $\sigma^2$
$=\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{d}_{\text{i}}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{d}_{\text{i}}\Big)^2$
$=\Big(\frac{1}{70}\times102\Big)-\Big(\frac{1}{70}\times(-38)\Big)^2$
$=1.457-0.295$
$=1.162\text{gm}$
Standard deviation, $\sigma=\sqrt{\text{Variance}}=\sqrt{1.162}=1.08\text{gm}$ View full question & answer→Question 715 Marks
Find the standard deviation for the following data:
|
$x$
|
$3$
|
$8$
|
$13$
|
$18$
|
$23$
|
|
$f$
|
$7$
|
$10$
|
$15$
|
$10$
|
$6$
|
Answer
| $x$ |
$f$ |
$fx$ |
$x-$mean |
(x-mean)$^2$ |
f(x-mean)$^2$ |
| $3$ |
$7$ |
$21$ |
$-9.79$ |
$95.88$ |
$671.13$ |
| $8$ |
$10$ |
$80$ |
$-4.79$ |
$22.96$ |
$229.96$ |
| $13$ |
$15$ |
$195$ |
$0.21$ |
$0.04$ |
$0.65$ |
| $18$ |
$10$ |
$180$ |
$5.21$ |
$27.13$ |
$271.26$ |
| $23$ |
$6$ |
$138$ |
$10.21$ |
$104.21$ |
$625.26$ |
| |
$48$ |
$614$ |
|
|
$1797.92$ |
Here, N = 48, and $\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=614$
$\overline{\text{x}}=\frac{1}{\text{N}}\big(\sum\text{f}_{\text{i}}\text{x}_{\text{i}}\big)=\frac{614}{48}=12.79$
$\sum\text{f}_{\text{i}}(\text{x}_{\text{i}}-\overline{\text{x}})^2=1797.92$
$\therefore\text{Var}(\text{x})=\frac{1}{\text{N}}\Big[\sum\text{f}_{\text{i}}(\text{x}_{\text{i}}-\overline{\text{x}})^2\Big]=\frac{1797.92}{48}=37.46$
S.D. $=\sqrt{\text{var}(\text{x})}=\sqrt{37.496}=6.12$ View full question & answer→Question 725 Marks
The variance of 15 bservation is 4. If each observation is increased by 9, find the variance of the resulting observation.
AnswerWe have, n = 15, and $\sigma^2=4$
Now each observation is increasedd by 9.
Suposs X = x +9 be the new data.
$\therefore\overline{\text{X}}=\frac{1}{15}\sum(\text{x}_\text{i}+9)=\Big(\frac{1}{15}\times\sum\text{x}_\text{i}\Big)+9=\overline{\text{x}}+9$
$\Rightarrow\sum{\text{X}_\text{i}}^{2}=\sum(\text{x}_\text{i}+9)^2=\sum{\text{x}_\text{i}}^2+\sum18\text{x}_\text{i}+\sum9^2$
Since, $\sigma^2=5$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=4$
Now, for the new data:
$\sigma^2=\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=\frac{1}{15}\big(\sum{\text{x}_\text{i}}^2+\sum18\text{x}_\text{i}+\sum9^2\big)-\big(\overline{\text{x}}+9\big)^2$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2+\frac{1}{15}\sum18\text{x}_\text{i}+\frac{1}{15}\sum9^2-(9)^2-(18\overline{\text{x}})-(\overline{\text{x}})^2$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[\frac{1}{15}\sum18\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\sum9^2-(9)^2\Big]$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[18\times\frac{1}{15}\sum\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\times15\times(9)^2-(9)^2\Big]$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2$
$=4$
View full question & answer→Question 735 Marks
For a group of $200$ candidates, the mean and standard deviations of scores were found to be $40$ and $15$ respectively. Later on it was discovered that the scores of $43$ and $35$ were misread as $34$ and $53$ respectively. Find the correct mean and standard deviation.
AnswerWe have,
$\text{n}=200,\overline{\text{X}}=40,\sigma=15.$
$\therefore\overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}=\overline{\text{X}}=200\times40=8000.$
Corrected $\sum\text{x}_\text{i}$ = Incorrected $\sum\text{x}_\text{i}$ - (sum of incorrect values) + (sum of correct values)
= 8000 - 34 - 53 + 43 + 35 = 7991
$\therefore\text{Corrected}\ \text{mean}=\frac{\text{corrected}\sum\text{x}_\text{i}}{\text{n}}=\frac{7991}{200}=39.955$
Now $\sigma=15$
$\Rightarrow15^2=\frac{1}{200}\Big(\sum{\text{x}_\text{i}}^2\Big)-\Big(\frac{1}{200}\sum\text{x}_\text{i}\Big)^2$
$\Rightarrow255=\frac{1}{200}\Big(\sum{\text{x}_\text{i}}^2\Big)-\Big(\frac{8000}{200}\Big)^2$
$\Rightarrow255=\frac{1}{200}\Big(\sum{\text{x}_\text{i}}^2\Big)-1600$
$\sum{\text{x}_\text{i}}^2=200\times1825=365000$
Incorrect $\sum{\text{x}_\text{i}}^2=365000$
corrected $\sum{\text{x}_\text{i}}^2$ = $ \big(\text{incorrected}\sum{\text{x}_\text{i}}^2\big)$ - (sum of squers of incorrect values) + (sum of squers of correct values)
$= 365000 - (34)^2 - 53^2+ (43)^2+ 35^2 = 364109$
so, Corrected $\sigma=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\Big(\frac{1}{\text{n}}\sum\text{x}_\text{i}}\Big)^2=\sqrt{\frac{364109}{200}-\Big(\frac{7991}{200}}\Big)^2$
$=\sqrt{1820.545-1596.402}=14.97$
View full question & answer→