Download App

Maths Solve the Following Question.(4 Marks)

Statistics · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 44 questions.

Questions

Solve the Following Question.(4 Marks)

Take a timed test

44 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 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
$5$
$6$
$12$
$14$
$26$
$12$
$16$
$9$
Answer
Converting 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 24 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 34 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.
Answer
We 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 44 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 54 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 64 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
Answer
We 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 74 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 84 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 94 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 104 Marks
calculate the mean deviation from the mean for the following data:
$36, 72, 46, 42, 60, 45, 53, 46, 51, 49$
Answer
Let ${\overline{\text{x}}}$ 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 114 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
$8$
$10$
$15$
$25$
$20$
$18$
$9$
$5$
Answer
Computation 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 124 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
 
 
  1. Which firm A or B pays out larger amount as weekly wages?
  2. Which firm A or B has greater variability in individual wages?
Answer
Total wagas paid by firm A = (Averge wages) × (Number of employees)
= 52.5 × 587 = Rs 30817.50
Total wagas paid by firm A = (Averge wages) × (Number of employees)
= 47.5 × 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 134 Marks
Calculate the mean deviation of the following income groups of five and seven members from their medians:
$I$
Income in $₹$
$II$
Income in $₹$
$4000$
$3800$
$4200$ $4000$
$4400$ $4200$
$4600$ $4400$
$4800$ $4600$
  $4800$
  $5800$
Answer
Arrange 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 144 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
$5$
$8$
$15$
$16$
$6$
Answer
M.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 154 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.
$38, 70, 48, 34, 63, 42, 55, 44, 53, 47$
Answer
Let $\overline{\text{x}}$ 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 164 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
Answer
We 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 174 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 184 Marks
Calculate the mean deviation about the median of the following observation:
$34, 66, 30, 38, 44, 50, 40, 60, 42, 51$
Answer
Formula 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 194 Marks
Calculate the mean deviation about the median of the following observation:
$38, 70, 48, 34, 42, 55, 63, 46, 54, 44$
Answer
Formula 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 204 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 214 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 224 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?
Answer
In 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 234 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 244 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 254 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 264 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 274 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 284 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
Answer
We 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 294 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 304 Marks
Calculate the mean deviation about the median of the following observation:
$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$
Answer
Formula 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 314 Marks
The variance of 20 observation is 5. If each observation is multiplied by 2, find the variance of the resulting observation.
Answer
We 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 324 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 334 Marks
calculate the mean deviation from the mean for the following data:
$57, 64, 43, 67, 49, 59, 44, 47, 61, 59$
Answer
First 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 344 Marks
Calculate the mean deviation about the median of the following observation:
$38, 70, 48, 34, 63, 42, 55, 44, 53, 47$
Answer
Formula 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 354 Marks
Calculate the mean deviation about the median of the following observation:
$3011, 2780, 3020, 2354, 3541, 4150, 5000$
Answer
Formula 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 364 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 374 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 384 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 394 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 404 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 414 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 $S.D.$
Answer
We 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$)^2$ 
$\Rightarrow $ Corrected variance $=\frac{161701}{100}-\Big(\frac{3990}{100}\Big)^2$
$\Rightarrow $ Corrected variance $=\frac{161701\times100-(3990)^2}{(100)^2}$
$\Rightarrow $ Corrected variance $=\frac{16170100-15920100}{10000}=25$
$\therefore$ Correctes standard deviation $=\sqrt{25}=5$
View full question & answer→
Question 424 Marks
calculate the mean deviation about median of the following frequency distribution:
$x_i$
$5$
$7$
$9$
$11$
$13$
$15$
$17$
$f_i$
$2$
$4$
$6$
$8$
$10$
$12$
$8$
Answer
We 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 434 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?
Answer
In 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 444 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→
Generate Paper FreeAll Statistics topics