## NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.3) Exercise 15.3

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## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

1. From the data given below, which group is more variable, A or B ?

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |

Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |

Ans: First we have to find the standard deviation of group A.

Standard deviation of A is calculate as follows,

Marks | Group A \[\left( {{f_i}} \right)\] | Mid-point \[\left( {{x_i}} \right)\] | \[{y_i} = \dfrac{{{x_i} - 45}}{{10}}\] | \[{y_i}^2\] | \[{f_i}{y_i}\] | \[{f_i}{y_i}^2\] |

10-20 | 9 | 15 | \[ - 3\] | 9 | \[ - 27\] | 81 |

20-30 | 17 | 25 | \[ - 2\] | 4 | \[ - 34\] | 68 |

30-40 | 32 | 35 | \[ - 1\] | 1 | \[ - 32\] | 32 |

40-50 | 33 | 45 | 0 | 0 | 0 | 0 |

50-60 | 40 | 55 | 1 | 1 | 40 | 40 |

60-70 | 10 | 65 | 2 | 4 | 20 | 40 |

70-80 | 9 | 75 | 3 | 9 | 27 | 81 |

150 | \[ - 6\] | 342 |

Here,

\[h = 10\]

\[N = 150\]

\[A = 45\]

\[{\text{Mean = }}A + \dfrac{{\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} }}{N} \times h\]

\[{\text{Mean = }}45 + \dfrac{{\left( { - 6} \right)}}{{150}} \times 10\]

\[{\text{Mean = }}45 - 0.4\]

\[{\text{Mean = }}44.6\]

Calculating standard deviation,

\[{\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} - {{\left( {\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} } \right)}^2}} \right]\]

\[{\sigma ^2} = \dfrac{{100}}{{22500}}\left[ {150 \times 342 - {{\left( { - 6} \right)}^2}} \right]\]

\[{\sigma ^2} = \dfrac{1}{{225}}\left( {51264} \right)\]

\[\sigma = \sqrt {227.84}\]

\[\sigma = 15.09\]

\[\sigma = \sqrt {227.84}\]

\[\sigma = 15.09\]

Now we have to find the standard deviation of group B.

Standard deviation of B is calculate as follows,

Marks | Group A \[\left( {{f_i}} \right)\] | Mid-point \[\left( {{x_i}} \right)\] | \[{y_i} = \dfrac{{{x_1} - 45}}{{10}}\] | \[{y_i}^2\] | \[{f_i}{y_i}\] | \[{f_i}{y_i}^2\] |

10-20 | 10 | 15 | \[ - 3\] | 9 | \[ - 30\] | 90 |

20-30 | 20 | 25 | \[ - 2\] | 4 | \[ - 40\] | 80 |

30-40 | 30 | 35 | \[ - 1\] | 1 | \[ - 30\] | 30 |

40-50 | 25 | 45 | 0 | 0 | 0 | 0 |

50-60 | 43 | 55 | 1 | 1 | 43 | 43 |

60-70 | 15 | 65 | 2 | 4 | 30 | 60 |

70-80 | 7 | 75 | 3 | 9 | 21 | 63 |

150 | \[ - 6\] | 366 |

\[{\text{Mean = }}A + \dfrac{{\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} }}{N} \times h\]

\[{\text{Mean = }}45 + \dfrac{{\left( { - 6} \right)}}{{150}} \times 10\]

\[{\text{Mean = }}45 - 0.4\]

\[{\text{Mean = }}44.6\]

Calculating standard deviation,

\[{\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} - {{\left( {\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} } \right)}^2}} \right]\]

\[{\sigma ^2} = \dfrac{{100}}{{22500}}\left[ {150 \times 366 - {{\left( { - 6} \right)}^2}} \right]\]

\[{\sigma ^2} = \dfrac{1}{{225}}\left( {54846} \right)\]

\[{\sigma ^2} = 243.84\]

\[\sigma = \sqrt {243.84} \]

\[\sigma = 15.61\]

As the mean of both the groups is same, therefore, the group with greater standard deviation is more variable.

Therefore, group B is more variable.

2. From the prices of shares X and Y below, find out which is more stable.

X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |

Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |

Ans: To compare the stability of both shares we have to calculate the standard deviation of both shares.

Coefficient of variance i.e. C.V. of share X can be calculated as,

The prices of shares X are,

\[35,54,52,53,56,58,52,50,51,49\]

Total number of observation is 10

Mean,

\[\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{x_i}} \]

\[\overline x = \dfrac{1}{{10}} \times 510\]

\[\overline x = 51\]

\[{x_i}\] | \[\left( {{x_i} - \overline x } \right)\] | \[{\left( {{x_i} - \overline x } \right)^2}\] |

35 | \[ - 16\] | 256 |

54 | 3 | 9 |

52 | 1 | 1 |

53 | 2 | 4 |

56 | 5 | 25 |

58 | 7 | 49 |

52 | 1 | 1 |

50 | \[ - 1\] | 1 |

51 | 0 | 0 |

49 | \[ - 2\] | 4 |

350 |

Variance of shares of X can be calculated as,

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \overline x } \right)}^2}} \]

\[{\sigma ^2} = \dfrac{1}{{10}} \times 350\]

\[{\sigma ^2} = 35\]

\[\sigma = \sqrt {35} \]

\[\sigma = 5.91\]

\[C.V.{\text{ of shares X }} = \dfrac{\sigma }{{\overline x }} \times 100\]

\[C.V.{\text{ of shares X}} = \dfrac{{5.91}}{{51}} \times 100\]

\[C.V.{\text{ of shares X}} = 11.58\]

Coefficient of variance i.e. C.V. of share Y can be calculated as,

The prices of shares Y are,

\[108,107,105,105,106,107,104,103,104,101\]

Total number of observation is 10

Mean,

\[\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{x_i}} \]

\[\overline x = \dfrac{1}{{10}} \times 1050\]

\[\overline x = 105\]

\[{y_i}\] | \[\left( {{y_i} - \overline y } \right)\] | \[{\left( {{y_i} - \overline y } \right)^2}\] |

108 | 3 | 9 |

107 | 2 | 4 |

105 | 0 | 0 |

105 | 0 | 0 |

106 | 1 | 1 |

107 | 2 | 4 |

104 | \[ - 1\] | 1 |

103 | \[ - 2\] | 4 |

104 | \[ - 1\] | 1 |

101 | \[ - 4\] | 16 |

40 |

Variance of shares of Y can be calculated as,

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{y_i} - \overline y } \right)}^2}} \]

\[{\sigma ^2} = \dfrac{1}{{10}} \times 40\]

\[{\sigma ^2} = 4\]

\[\sigma = \sqrt 4 \]

\[\sigma = 2\]

\[C.V.{\text{ of shares X }} = \dfrac{\sigma }{{\overline x }} \times 100\]

\[C.V.{\text{ of shares X}} = \dfrac{4}{{105}} \times 100\]

\[C.V.{\text{ of shares X}} = 1.9\]

On comparing the \[C.V.\] of both X and Y,

\[C.V.\]of X is greater than \[C.V.\] of Y.

Therefore, the price of Y is more stable than the price of X.

3. An analysis of monthly 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 wages earners | 586 | 648 |

Mean of monthly wages | Rs. 5253 | Rs. 5253 |

Variance of the distribution of wages | 100 | 121 |

(i) which firm A or B pays larger amounts as monthly wages ?

Ans: Monthly wages of firm A is Rs. 5253.

Total number of wage earners in firm A is 586.

Therefore, total amount paid in firm A is \[{\text{Rs}}{\text{. }}5253 \times 586 = {\text{ Rs}}{\text{. }}3078258\] .

Monthly wages of firm B is Rs. 5253.

Total number of wage earners in firm B is 648.

Therefore, total amount paid in firm B is \[{\text{Rs}}{\text{. }}5253 \times 648 = {\text{ Rs}}{\text{. }}3403944\] .

Therefore, firm B pays a larger amount as monthly wages.

(ii) Which firm A or B, shows greater variability in individual wages ?

Ans: Variance of distribution \[\left( {{\sigma ^2}} \right)\] of wages of firm A is 100.

So, the standard deviation of the wages in firm A is

\[{\sigma ^2} = 100\]

\[\sigma = \sqrt {100} \]

\[\sigma = 10\]

Variance of distribution \[\left( {{\sigma ^2}} \right)\] of wages of firm B is 121.

So, the standard deviation of the wages in firm B is

\[{\sigma ^2} = 121\]

\[\sigma = \sqrt {121} \]

\[\sigma = 11\]

Since, the standard deviation of firm B is greater than firm B,

Therefore, firm B has greater variability in the individual wages.

4. The following is the record of goals scored by team A in a football session:

No. of goals scores | 0 | 1 | 2 | 3 | 4 |

No. of matches | 1 | 9 | 7 | 5 | 3 |

For team B, the mean number of goals scored per match was 2 with a standard deviation \[1.25\] goals. Find which team may be considered more consistent?

Ans: We have to compare team A and B, to see which team is more consistent.

For that we will calculate standard deviation of team A.

Calculation of Standard Deviation of Team A:

No. goals scored\[\left( {{x_i}} \right)\] | No of matches\[\left( {{f_i}} \right)\] | \[{f_i}{x_i}\] | \[{x_i}^2\] | \[{f_i}{x_i}^2\] |

0 | 1 | 0 | 0 | 0 |

1 | 9 | 9 | 1 | 9 |

2 | 7 | 14 | 4 | 28 |

3 | 5 | 15 | 9 | 45 |

4 | 3 | 12 | 16 | 48 |

25 | 130 |

Mean,

\[\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{f_i}{x_i}} \]

\[\overline x = \dfrac{1}{{25}} \times 50\]

\[\overline x = 2\]

Standard deviation of A is,

\[\sigma = \dfrac{1}{N}\sqrt {N\sum {{x_i}{f_i}} - {{\left( {\sum {{x_i}{f_i}} } \right)}^2}} \]

\[\sigma = \dfrac{1}{{25}}\sqrt {25 \times 130 - {{\left( {50} \right)}^2}} \]

\[\sigma = \dfrac{1}{{25}}\sqrt {750} \]

\[\sigma = 1.09\]

Standard deviation of team A is \[1.09\] .

Standard deviation of team B is \[1.25\] .

Since the mean of both teams is the same i.e. 2.

Standard deviation of team B is greater than team A.

Therefore, team A is more consistent.

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

\[\sum\limits_{i = 1}^{50} {{x_i}} = 212,\sum\limits_{i = 1}^{50} {{x_i}^2} = 902.8,\sum\limits_{i = 1}^{50} {{y_i}} = 261,\sum\limits_{i = 1}^{50} {{y_i}^2} = 1457.6\]

Which is more varying, the length or weight?

Ans: Calculating standard deviation and coefficient of length.

\[\sum\limits_{i = 1}^{50} {{x_i}} = 212,\sum\limits_{i = 1}^{50} {{x_i}^2} = 902.8\]

Here, we have given that,

\[N = 50\]

Therefore, mean will be

\[\overline x = \dfrac{{\sum\limits_{i = 1}^{50} {{x_i}} }}{N}\]

\[\overline x = \dfrac{{212}}{{50}}\]

\[\overline x = 4.24\]

Now, calculating variance,

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - \overline x } \right)}^2}} \]

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {\left( {{x_i}^2 + {{\overline x }^2} - 2{x_i}.\overline x } \right)} \]

\[{\sigma ^2} = \dfrac{1}{N}\left( {\sum\limits_{i = 1}^{50} {{x_i}^2} + \sum\limits_{i = 1}^{50} {{{\left( {\overline x } \right)}^2}} - \sum\limits_{i = 1}^{50} {2{x_i}.\overline x } } \right)\]

\[{\sigma ^2} = \dfrac{1}{{50}}\left( {902.8 + 17.97 - 2 \times 4.24 \times 212} \right)\]

\[{\sigma ^2} = \dfrac{1}{{50}}\left( {3.54} \right)\]

\[{\sigma ^2} = 0.07\]

\[\sigma = \sqrt {0.07} \]

\[\sigma = 0.26\]

So, the standard deviation of length is \[0.26\] .

\[{\text{C}}{\text{.V}}{\text{. = }}\dfrac{\sigma }{{\overline x }} \times 100\]

\[{\text{C}}{\text{.V}}{\text{. = }}\dfrac{{0.26}}{{4.24}} \times 100\]

\[{\text{C}}{\text{.V}}{\text{. = }}6.13\]

So, the coefficient of variance of length is \[6.13\] .

Calculating standard deviation and coefficient of variance of weigth.

\[\sum\limits_{i = 1}^{50} {{y_i}} = 261,\sum\limits_{i = 1}^{50} {{y_i}^2} = 1457.6\]

Here, we have given that,

\[N = 50\]

Therefore, mean will be

\[\overline y = \dfrac{{\sum\limits_{i = 1}^{50} {{y_i}} }}{N}\]

\[\overline y = \dfrac{{261}}{{50}}\]

\[\overline y = 5.22\]

Now, calculating variance,

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - \overline y } \right)}^2}} \]

\[{\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {\left( {{y_i}^2 + {{\overline y }^2} - 2{y_i}.\overline y } \right)} \]

\[{\sigma ^2} = \dfrac{1}{N}\left( {\sum\limits_{i = 1}^{50} {{y_i}^2} + \sum\limits_{i = 1}^{50} {{{\left( {\overline y } \right)}^2}} - \sum\limits_{i = 1}^{50} {2{y_i}.\overline y } } \right)\]

\[{\sigma ^2} = \dfrac{1}{{50}}\left( {1457.6 + 27.24 - 2 \times 5.22 \times 261} \right)\]

\[{\sigma ^2} = \dfrac{1}{{50}}\left( {94.76} \right)\]

\[{\sigma ^2} = 1.89\]

\[\sigma = \sqrt {1.89} \]

\[\sigma = 1.37\]

So, the standard deviation of weight is \[1.37\] .

\[{\text{C}}{\text{.V}}{\text{. = }}\dfrac{\sigma }{{\overline y }} \times 100\]

\[{\text{C}}{\text{.V}}{\text{. = }}\dfrac{{1.37}}{{5.22}} \times 100\]

\[{\text{C}}{\text{.V}}{\text{. = 26}}{\text{.24}}\]

So, the coefficient of variance of weight is \[{\text{26}}{\text{.24}}\] .

Since the coefficient of variance of weight is greater than the coefficient of variance of length.

Therefore, weight varies more than length.

## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

Opting for the NCERT solutions for Ex 15.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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