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Maths 5 Marks Questions

Statistics · Jharkhand Board (JAC) STD 11 Science (Jharkhand - English Medium). 73 questions.

Questions · Page 2 of 2

5 Marks Questions

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.
$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$
Answer
Let $\overline{\text{x}}$ 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.
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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 $\overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
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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.
Answer
Let 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.
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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.
Answer
We 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,
$\text{Incorrect}\sum{\text{x}_\text{i}}^2=40900$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-21^2-21^2-18^2$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-1206$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=39694$
$\therefore\text{Corrected}\ \text{mean}=\frac{1940}{97}=20$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{1}{97}\big(\text{Corrected}\sum{\text{x}_\text{i}}^2\big)-\big(\text{Corrected}\ \text{mean}\big)^2$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{39694}{97}-(20)^2=409.22-400=9.22$
$​​\therefore\text{Corrected}\ \text{standard}\ \text{deviation}=\sqrt{9.22}=3.04$
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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 $h = 10$
$\text{Mean}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\text{Mean}=35+10\Big(\frac{-98}{70}\Big)=-21$
$\text{Var}\big(\text{X}\big)=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\bigg\{\Big(\frac{1}{70}\times250\Big)-\Big(\frac{1}{70}\times(-98)\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\big\{3.57-1.96\big\}=161$
$\text{SD}=\sqrt{\text{Var}(\text{X})}=\sqrt{161}=12.7$
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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,
$\sum|\text{x}_\text{i}-14|=\sum\text{d}_\text{i}=28$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{12}[28]=2.33$
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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.
Answer
Mean, $\overline{\text{x}}=\frac{\sum\text{f}_{\text{i}}\text{x}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{10078}{2600}=3.88$
$x_i$
$f_i$
$f_ix_i$
$\text{x}_{\text{i}}-\overline{\text{X}}$
$\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
$\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
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, $\sigma^2=\frac{\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2​​}{{\text{N}}}=\frac{9448.848}{2600}=3.63$
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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.
Answer
Mean = 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 $\sigma=\sqrt{\frac{164938}{100}-(39.3)^2}=10.24$
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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 $=\sqrt{8.12}=2.85$
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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?
Answer
For 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.
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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 $=\sqrt{239.96}=15.49$
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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.
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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
Answer
Let'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.
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Question 645 Marks
The lengths $($in $cm)$ of $10$ rods in a shop are given below:
$40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2$
  1. Find mean deviation from median
  2. Find mean deviation from the mean also.
Answer
First 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 $167.4$ $166.4$
  1. $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{167.4}{10}=16.74$
  2. $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{166.4}{10}=16.64$
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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?
Answer
Factor 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.
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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, $\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2=12308.57$
$\therefore\text{ver}(\text{x})=\frac{1}{\text{N}}\Big[\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2\Big]\\=\frac{12308.57}{70}=175.84$
$\text{S.D.}=\sqrt{\text{ver}(\text{x})}=\sqrt{175.84}=13.26$
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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.
Answer
Let 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.
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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,
$\text{Mean}=\frac{\sum\text{f}_{\text{i}}\text{y}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{66.5}{16}=4.16$
$\text{Variance}=\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}\Big)^2$
$\text{Variance}=\frac{1}{16}\times340.25-\Big(\frac{1}{16}\times66.5\Big)^2$
$\text{Variance}=21.26-17.22=4.04$
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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?
Answer
We 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 $\sigma=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\Big(\frac{1}{\text{n}}\sum\text{x}_\text{i}}\Big)^2=\sqrt{\frac{161701}{100}-\Big(\frac{3990}{100}}\Big)^2$
$=\sqrt{1617.01-1592.01}=5$
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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, $\sigma=\sqrt{\text{Variance}}=\sqrt{1.162}=1.08\text{gm}$
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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. $=\sqrt{\text{var}(\text{x})}=\sqrt{37.496}=6.12$
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Question 725 Marks
The variance of 15 bservation is 4. If each observation is increased by 9, find the variance of the resulting observation.
Answer
We 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:
$\sigma^2=\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=\frac{1}{15}\big(\sum{\text{x}_\text{i}}^2+\sum18\text{x}_\text{i}+\sum9^2\big)-\big(\overline{\text{x}}+9\big)^2$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2+\frac{1}{15}\sum18\text{x}_\text{i}+\frac{1}{15}\sum9^2-(9)^2-(18\overline{\text{x}})-(\overline{\text{x}})^2$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[\frac{1}{15}\sum18\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\sum9^2-(9)^2\Big]$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[18\times\frac{1}{15}\sum\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\times15\times(9)^2-(9)^2\Big]$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2$
$=4$
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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.
Answer
We 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 $\sigma=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\Big(\frac{1}{\text{n}}\sum\text{x}_\text{i}}\Big)^2=\sqrt{\frac{364109}{200}-\Big(\frac{7991}{200}}\Big)^2$
$=\sqrt{1820.545-1596.402}=14.97$
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