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Class 11 RD Sharma Solutions- Chapter 32 Statistics – Exercise 32.6

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  • Last Updated : 21 Feb, 2021
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Question 1. Calculate the mean and S.D. for the following data:

Expenditure (in ₹): 0-10 10-20 20-30 30-40 40-50
Frequency: 14 13 27 21 15

Solution:

CI   f x u = (x – A)/h   fu u2 fu2
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

Given: 

Number of observations, N = 90 and A = 25

\sum f_iu_i =10\\ \sum f_iu_i^2  =150    

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i)

\overline{x} = 25 + 10(10/90) = 26.11 

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 10[(150/90) – (10/90)2]

= 165.4 

Standard Deviation = √var(x) = √165.4 = 12.86  

Question 2. 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

Solution:

CI f   x u = (x – A)/h f × u u2 fu2
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
  300   137   665

Given: N = 300 and A =105

\sum f_iu_i =137\\ \sum f_iu_i^2  =665

h = 30

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i)

\overline{x} = 105 + 30(137/300) = 118.7

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 900[(665/300) – (137/300)2

= 1807.31 

Standard Deviation = √var(x) = √1807.31 = 42.51

Question 3. 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

Solution:

CI f   x u = (x – A)/h f × u u2 fu2
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

Given:  N = 79 and A =35

\sum f_iu_i =-71\\ \sum f_iu_i^2  =305

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i)

\overline{x} = 35 + 10(-71/79) = 26.01

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 100[(305/79) – (-71/79)2

= 305.30

Standard Deviation  = √var(x) = √305.30 = 17.47

Question 4. 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 is 40. Find the correct mean and S.D.

Solution:

According to question, we have, 

n = 100 , \overline{x} = 40 , \sigma = 5.1 \\ \overline{x} = \frac{1}{n}\sum x_i \\ \sum x_i = n\overline{x} = 4000 \\ Incorrect \sum x_i = 4000

And, also 

\sigma = 5.1 \\ \sigma^2 = 26.01 \\ \frac{1}{n}\sum x_i^2 - Mean^2 = 26.01 \\ \frac{1}{100}\sum x_i^2 - 1600 = 26.01 \\

\sum x_i^2 = 1626.01 x 100 

Incorrected \sum x_i^2 = 162601

On replacing the incorrect observation of 50 by 40, we get,

Incorrect \sum x_i  = 4000 

Corrected \sum x_i = 4000 – 50 + 40 = 3990

Incorrected \sum x_i^2 = 162601 

Corrected \sum x_i^2 = 162601 – 502 + 402 = 161701

Now, we have,

Corrected Mean = 39.90

Corrected Variance = (1/100)(Corrected \sum x_i^2) – (Corrected mean)2 

 = \frac{161701}{100} - (\frac{3990}{100})^2 \\ = \frac{161701 * 100 - (3990)^2}{100^2} \\ = \frac{16170100-15920100}{10000}

= 25

Corrected standard deviation = √25 = 5

Question 5. 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

Solution:

CI Freq Mid Value ui fiui fiui2
31 – 35 2 33 -4 -8 32
36 – 40 3 38 -3 -9 27
41 – 45 8 43 -2 -16 32
46 – 50 12 48 -1 -12 12
51 – 55 16 53 0 0 0
56 – 60 5 58 1 5 5
61 – 65 2 63 2 4 8
66 – 70 2 68 3 6 18
  N = 50     Total = – 30 Total = 134

Now, using the given values, we have

Mean = 53 + 5 x (-30/50) 

= 50

Variance = 25 x ((134/50) – (9/25)

= 58

Standard Deviation = √58 

= 7.62

Question 6. Find the mean and variance of frequency distribution given below:

xi 1 ≤ x < 3 3 ≤ x < 5 5 ≤ x < 7 7 ≤ x < 9
fi 6 4 5 1

Solution:

The data can be converted to a continuous frequency distribution by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each of the class interval. 

Class Interval fi xi ui fiui ui2 fiui2
1 – 2 6 1.5 -4 -24 16 96
3 – 4 4 3.5 -2 -8 4 16
5 – 6 5 5.5 0 0 0 0
7 – 8 1 7.5 2 2 4 4
  N = 16     Total = -30   Total = 116

Given: N = 16 and A = 5.5

\sum f_iu_i =-30\\ \sum f_iu_i^2  =116   and h=1

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i)

\overline{x} = 5.5 + 1((1/6) x (-30))

= 3.625 

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 1 [((1/16) x 116) – ((1/16) x (-30)2

= 3.74

Question 7. 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

Calculate mean, variance, and standard deviation.

Solution:

CI xi fi ui fiui fiui2
200 – 201 200.5 13 -15 -19.5 29.25
201 – 202 201.5 27 -1 -27 27
202 – 203 202.5 18 -0.5 -9 4.5
203 – 204 203.5 10 0 0 0
204 – 205 204.5 1 0.5 0.5 0.25
205 – 206 205.5 1 1 1 1
    N = 70   Total = – 54 Total = 62

Now, using the given values, we have

Mean = 203.5 + 2 x (-54/70)

= 201.9

Variance = 4 x (62/70) – (-54/70)

= 0.98

Standard Deviation = √0.98

= 0.099

Question 8. 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.

Solution:

Mean = 40

Standard Deviation = 10

n = 100

\sum x_i = 40 * 100 = 4000

Corrected Sum = 4000 – 30 +70 + 3 + 27 = 3930

Corrected mean = 39.3

Variance = 100

100 = \frac{\sum x_i^2}{100} -(40)^2

Incorrect \sum x_i^2 = 170000 

So, Corrected \sum x_i^2 = Incorrect \sum x_i^2 – (Sum of squares of incorrect values) + 

                                                                                    (Sum of squares of corrected values) 

Corrected\sum x_i^2 = 170000 – (900 + 4900) + (9+729)  

= 164938

Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{164938}{100}-(39.3)^2} \\

 = 10.24

Question 9. While calculating the mean and variance of 10 reading, 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 variance. 

Solution:

Mean = 45

Variance = 16

n = 10

\sum x_i = 450

So, Corrected Sum = 450 – 52 + 25 = 423

Corrected mean = 42.3

Variance = 16

116 = \frac{\sum x_i^2}{10} -(45)^2 \\ Incorrect \sum x_i^2 = 20410

Corrected \sum x_i^2 = Incorrect  \sum x_i^2– (Sum of squares of incorrect values) + 

                                                                             (Sum of squares of corrected values) 

Corrected \sum x_i^2  = 20410 – 2704 + 625 = 18331

Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{18331}{10}-(42.3)^2}

= 6.62

So, Corrected variance = 6.62 * 6.62 = 43.82

Question 10. Calculate mean, variance, and standard deviation of the following frequency distribution:

Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 11 29 18 4 5 3

Solution:

CI xi fi ui fiui fiui2
0-10 5 11 -3 -33 99
10-20 15 29 -2 -58 116
20-30 25 18 -1 -18 18
30-40 35 4 0 0 0
40-50 45 5 1 5 5
50-60 55 3 2 6 12
    N = 70   Total = – 98 Total = 250

Given:

Number of observations, N = 70 and A = 35

\sum f_iu_i =-98\\ \sum f_iu_i^2  =250

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i)

\overline{x} = 35 + 10(-98/70) = -21

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 100[(1/70) x 250 – (1/70) x (-98)2]

= 161 

Standard Deviation = √var(x) 

= √161

= 12.7


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