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

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

Expenditure (in ₹):0-1010-2020-3030-4040-50
Frequency:1413272115

Solution:

CI  fxu = (x – A)/h  fuu2fu2
0 – 10145-2-28456
10 – 201315-1-13113
20 – 3027250000
30 – 402135121121
40 – 501545230460
 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:0-3030-6060-9090-120120-150150-180180-210
Frequency:9174382814424

Solution:

CIf  xu = (x – A)/hf × uu2fu2
0 – 30915-3-27981
30 – 601745-2-34468
60 – 904375-1-43143
90 – 120821050000
120 – 15081135181181
150 – 180441652884176
180 – 210241953729216
 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:0-1010-2020-3030-4040-5050-6060-7070-80
Frequency:1816151210521

Solution:

CIf  xu = (x – A)/hf × uu2fu2
0 – 10185-3-549162
10 – 201615-2-32464
20 – 301525-1-15115
30 – 4012350000
40 – 501045110110
50 – 60555210420
60 – 7026536918
70 – 80175441616
 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-interval31-3536-4041-4546-5051-5556-6061-6566-70
Frequency:2381216523

Solution:

CIFreqMid Valueuifiuifiui2
31 – 35233-4-832
36 – 40338-3-927
41 – 45843-2-1632
46 – 501248-1-1212
51 – 551653000
56 – 60558155
61 – 65263248
66 – 702683618
 N = 50  Total = – 30Total = 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:

xi1 ≤ x < 33 ≤ x < 55 ≤ x < 77 ≤ x < 9
fi6451

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 Intervalfixiuifiuiui2fiui2
1 – 261.5-4-241696
3 – 443.5-2-8416
5 – 655.50000
7 – 817.52244
 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)200-201201-202202-203203-204204-205205-206
Frequency1327181011

Calculate mean, variance, and standard deviation.

Solution:

CIxifiuifiuifiui2
200 – 201200.513-15-19.529.25
201 – 202201.527-1-2727
202 – 203202.518-0.5-94.5
203 – 204203.510000
204 – 205204.510.50.50.25
205 – 206205.51111
  N = 70 Total = – 54Total = 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:

Class0-1010-2020-3030-4040-5050-60
Frequency112918453

Solution:

CIxifiuifiuifiui2
0-10511-3-3399
10-201529-2-58116
20-302518-1-1818
30-40354000
40-50455155
50-605532612
  N = 70 Total = – 98Total = 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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