NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3

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Chapter 15 Ex.15.3 Question 1

From the data given below state 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 \(B\)

\(10\)

\(20\)

\(30\)

\(25\)

\(43\)

\(15\)

\(7\)

Solution

Standard deviation of Group \(A\) is calculated as follows.

Marks

Group \(A\)

\({f_i}\)

Mid-point \({x_i}\)

\({y_i} = \frac{{{x_i} - 45}}{{10}}\)

\({y_i}^2\)

\({f_i}{y_i}\)

\({f_i}{y_i}^2\)

\(9\)

\(10 - 20\)

\(15\)

\( - 3\)

\(9\)

\( - 27\)

\(81\)

\(20 - 30\)

\(17\)

\(25\)

\( - 2\)

\(4\)

\( - 34\)

\(68\)

\(30 - 40\)

\(32\)

\(35\)

\( - 1\)

\(1\)

\( - 32\)

\(32\)

\(40 - 50\)

\(33\)

\(45\)

\(0\)

\(0\)

\(0\)

\(0\)

\(50 - 60\)

\(40\)

\(55\)

\(1\)

\(1\)

\(40\)

\(40\)

\(60 - 70\)

\(10\)

\(65\)

\(2\)

\(4\)

\(20\)

\(40\)

\(70 - 80\)

\(9\)

\(75\)

\(3\)

\(9\)

\(27\)

\(81\)

 

\(150\)

 

 

 

\( - 6\)

\(342\)

Here, \(N = 150,\;h = 10,\;A = 45\)

Mean,

\[\begin{align}\bar x &= A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h\\&= 45 + \frac{{\left( { - 6} \right)}}{{150}} \times 10\\&= 45 - 0.4\\&= 44.6\end{align}\]

Variance,

\[\begin{align}\left( {{\sigma _1}^2} \right) &= \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i} - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{100}}{{22500}}\left[ {150 \times 342 - {{\left( { - 6} \right)}^2}} \right]\\&= \frac{1}{{225}} \times 51264\\&= 227.84\end{align}\]

Standard deviation,

\[\begin{align}\left( {{\sigma _1}} \right)&= \sqrt {227.84} \\&= 15.09\end{align}\]

Standard deviation of Group B is calculated as follows.

Marks

Group B

\({f_i}\)

Mid-point \({x_i}\)

\({y_i} = \frac{{{x_i} - 45}}{{10}}\)

\({y_i}^2\)

\({f_i}{y_i}\)

\({f_i}{y_i}^2\)

\(10 - 20\)

\(10\)

\(15\)

\( - 3\)

\(9\)

\( - 30\)

\(90\)

\(20 - 30\)

\(20\)

\(25\)

\( - 2\)

\(4\)

\( - 40\)

\(80\)

\(30 - 40\)

\(30\)

\(35\)

\( - 1\)

\(1\)

\( - 30\)

\(30\)

\(40 - 50\)

\(25\)

\(45\)

\(0\)

\(0\)

\(0\)

\(0\)

\(50 - 60\)

\(43\)

\(55\)

\(1\)

\(1\)

\(43\)

\(43\)

\(60 - 70\)

\(15\)

\(65\)

\(2\)

\(4\)

\(30\)

\(60\)

\(70 - 80\)

\(7\)

\(75\)

\(3\)

\(9\)

\(21\)

\(63\)

 

\(150\)

 

 

 

\( - 6\)

\(366\)

Mean,

\[\begin{align}\bar x &= A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h\\&= 45 + \frac{{\left( { - 6} \right)}}{{150}} \times 10\\&= 45 - 0.4\\&= 44.6\end{align}\]

Variance,

\[\begin{align}{\sigma _2}^2 &= \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i} - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]\\&= \frac{{100}}{{22500}}\left[ {150 \times 366 - {{\left( { - 6} \right)}^2}} \right]\\&= \frac{1}{{225}} \times 54864\\&= 243.84\end{align}\]

Standard deviation,

\(\begin{array}{c}\left( {{\sigma _2}} \right) = \sqrt {243.84} \\= 15.61\end{array}\)

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

Chapter 15 Ex.15.3 Question 2

From the prices of shares of \(X\) and \(Y\) below, find out which is more stable in value:

\(X\)

\(35\)

\(54\)

\(52\)

\(53\)

\(56\)

\(58\)

\(52\)

\(50\)

\(51\)

\(49\)

\(Y\)

\(108\)

\(107\)

\(105\)

\(105\)

\(106\)

\(107\)

\(104\)

\(103\)

\(104\)

\(101\)

Solution

The prices of the shares \(X\) are \(35,54,52,53,56,58,52,50,51,49\)

Here, the number of observations, \(N = 10\)

Mean,

\[\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^{10} {x_i} \\ &= \frac{1}{10} \times 510\\ &= 51\end{align}\]

The following table is obtained corresponding to shares \(X.\)

\({x_i}\)

\(\left( {{x_i} - \bar x} \right)\)

\({\left( {{x_i} - \bar x} \right)^2}\)

\(35\)

\( - 16\)

\(256\)

\(54\)

\(3\)

\(9\)

\(52\)

\(1\)

\(1\)

\(53\)

\(2\)

\(4\)

\(56\)

\(5\)

\(25\)

\(58\)

\(7\)

\(49\)

\(52\)

\(1\)

\(1\)

\(50\)

\(-1\)

\(1\)

\(51\)

\(0\)

\(0\)

\(49\)

\(-2\)

\(4\)

 

 

\(350\)

Variance,

\[\begin{align}\left( {{\sigma _1}^2} \right) &= \frac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \bar x} \right)}^2}} \\&= \frac{1}{{10}} \times 350\\&= 35\end{align}\]

Standard deviation,

\[\begin{align}\left( {{\sigma _1}} \right) &= \sqrt {35} \\&= 5.91\end{align}\]

\[\begin{align}  \text{C}.\text{V}.\left( \text{Shares X} \right)&=\frac{{{\sigma }_{1}}}{X}\times 100 \\ & =\frac{5.91}{51}\times 100 \\ & =11.58 \end{align}\]

The prices of the shares \(Y\) are \(108,107,105,105,106,107,104,103,104,101\)

Here, the number of observations, \(N = 10\)

Mean,

\[\begin{align}\bar y &= \frac{1}{N}\sum\limits_{i = 1}^{10} {{y_i}} \\&= \frac{1}{{10}} \times 1050\\&= 105\end{align}\]

The following table is obtained corresponding to shares \(Y.\)

\({y_i}\)

\(\left( {{y_i} - \bar y} \right)\)

\({\left( {{y_i} - \bar y} \right)^2}\)

\(108\)

\(3\)

\(9\)

\(107\)

\(2\)

\(4\)

\(105\)

\(0\)

\(0\)

\(105\)

\(0\)

\(0\)

\(106\)

\(1\)

\(1\)

\(107\)

\(2\)

\(4\)

\(104\)

\( - 1\)

\(1\)

\(103\)

\( - 2\)

\(4\)

\(104\)

\( - 1\)

\(1\)

\(101\)

\( - 4\)

\(16\)

 

 

\(40\)

Variance,

\[\begin{align}\left( {{\sigma _2}^2} \right) &= \frac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{y_i} - \bar y} \right)}^2}} \\&= \frac{1}{{10}} \times 40\\&= 4\end{align}\]

Standard deviation,

\(\left( {{\sigma _2}} \right) = \sqrt 4 = 2\)

\[\begin{align}  \text{C}.\text{V}.\left( \text{Shares Y} \right)&=\frac{{{\sigma }_{2}}}{X}\times 100 \\ & =\frac{2}{105}\times 100 \\ & =1.9 \end{align}\]

\(C.V\) of prices of shares \(X\) is greater than the \(C.V\) of prices of shares \(Y.\)

Thus, the prices of shares \(Y\) are more stable than the prices of shares \(X.\)

Chapter 15 Ex.15.3 Question 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

\({\rm{586}}\)

\(648\)

Mean of monthly wages

Rs. \(5253\)

Rs. \(5253\)

Variance of the distribution of wages

\(100\)

\(121\)

(i) Which firm \(A\) or \(B\) pays larger amount as monthly wages?

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

Solution

(i) Monthly wages of firm \(A=\) Rs \(5253\)

Number of wage earners in firm\( A=\)\({\rm{586}}\)

\(\therefore \)Total amount paid= Rs. \(5253 \times 586\)

Monthly wages of firm \(B= \)Rs \(5253\)

Number of wage earners in firm \(B=\)\(648\)

\(\therefore \)Total amount paid= Rs. \(5253 \times 648\)

Thus, firm \(B\) pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm \(A.\)

(ii)Variance of the distribution of wages in firm \(A\) \(\left( {\sigma _1^2} \right) = 100\)

\(\therefore \)Standard deviation of the distribution of wages in firm \(A\)\(\left( {{\sigma _1}} \right) = \sqrt {100} = 10\)

Variance of the distribution of wages in firm \(B\) \(\left( {\sigma _1^2} \right) = 121\)

\(\therefore \)Standard deviation of the distribution of wages in firm \(A\)\(\left( {{\sigma _1}} \right) = \sqrt {121} = 11\)

The mean of monthly wages of both the firms is same. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm \(B\) has greater variability in the individual wages.

Chapter 15 Ex.15.3 Question 4

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

No. of goals scored

\(0\)

\(1\)

\(2\)

\(3\)

\(4\)

No. of matches

\(1\)

\(9\)

\(7\)

\(5\)

\(3\)

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

Solution

The mean and standard deviation of goals scored by team A are calculated as follows.

No. of goals scored

No. of matches

\({f_i}{x_i}\)

\({x_i}^2\)

\({f_i}{x_i}^2\)

\(0\)

\(1\)

\(0\)

\(0\)

\(0\)

\(1\)

\(9\)

\(9\)

\(1\)

\(9\)

\(2\)

\(7\)

\(14\)

\(4\)

\(28\)

\(3\)

\(5\)

\(15\)

\(9\)

\(45\).

\(4\)

\(3\)

\(12\)

\(16\)

\(48\)

 

\(25\)

\(50\)

 

\(130\)

Mean,

\[\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} \\&= \frac{{50}}{{25}}\\&= 2\end{align}\]

Thus, the mean of both the teams is same.

\[\begin{align}\sigma &= \frac{1}{N}\sqrt {N\sum {{f_i}{x_i}^2 - {{\left( {{{\sum {{f_i}x} }_i}} \right)}^2}} } \\&= \frac{1}{{25}}\sqrt {25 \times 130 - {{\left( {50} \right)}^2}} \\&= \frac{1}{{25}}\sqrt {750} \\&= \frac{1}{{25}} \times 27.38\\&= 1.09\end{align}\]

The standard deviation of team B is \(1.25\) goals.

The average number of goals scored by both the teams is same i.e., \(2.\)

Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

Chapter 15 Ex.15.3 Question 5

The sum and sum of squares corresponding to length X (in cms) and weight y (in gms) of \({\rm{5}}0{\rm{ }}\)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 the weight?

Solution

\(\sum\limits_{i = 1}^{50} {{x_i} = 212} ,\sum\limits_{i = 1}^{50} {{x_i}^2 = 902.8} \)

Here, \(N = 50\)

Mean,

\[\begin{align}\bar x &= \frac{1}{N}\sum\limits_{i = 1}^{50} {{x_i}} \\&= \frac{{212}}{{50}}\\&= 4.24\end{align}\]

Variance,

\[\begin{align}\left( {{\sigma _1}^2} \right) &= \frac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - \bar x} \right)}^2}} \\&= \frac{1}{{50}}\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - 4.24} \right)}^2}} \\&= \frac{1}{{50}}\sum\limits_{i = 1}^{50} {\left[ {{x_i}^2 - 8.48{x_i} + 17.97} \right]} \\&= \frac{1}{{50}}\left[ {\sum\limits_{i = 1}^{50} {{x_i}^2} - 8.48\sum\limits_{i = 1}^{50} {{x_i}} + 17.97 \times 50} \right]\\A&= \frac{1}{{50}}\left[ {902.8 - 8.48 \times \left( {212} \right) + 898.5} \right]\\&= \frac{1}{{50}}\left[ {1801.3 - 1797.76} \right]\\&= \frac{1}{{50}} \times 3.54\\&= 0.07\end{align}\]

Standard variation \({\sigma _2}\left( {length} \right) = \sqrt {0.07} = 0.26\)

\[\begin{align}{\rm{C}}.{\rm{V}}\left( {{\rm{length}}} \right) &= \frac{{{\text{standard deviation}}}}{{mean}} \times 100\\&= \frac{{0.26}}{{4.24}} \times 100\\&= 6.13\end{align}\]

\(\sum\limits_{i = 1}^{50} {{y_i} = 261} ,\sum\limits_{i = 1}^{50} {{y_i}^2 = 1457.6} \)

Here, \(N = 50\)

Mean,

\[\begin{align}\bar y &= \frac{1}{N}\sum\limits_{i = 1}^{50} {{y_i}} \\&= \frac{{261}}{{50}}\\&= 5.22\end{align}\]

Variance,

\[\begin{align}\left( {{\sigma _2}^2} \right) &= \frac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - \bar y} \right)}^2}} \\&= \frac{1}{{50}}\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - 5.22} \right)}^2}} \\&= \frac{1}{{50}}\sum\limits_{i = 1}^{50} {\left[ {{y_i}^2 - 10.44{y_i} + 27.24} \right]} \\&= \frac{1}{{50}}\left[ {\sum\limits_{i = 1}^{50} {{y_i}^2} - 10.44\sum\limits_{i = 1}^{50} {{y_i}} + 27.24 \times 50} \right]\\&= \frac{1}{{50}}\left[ {1457.6 - 10.44 \times \left( {261} \right) + 1362} \right]\\&= \frac{1}{{50}}\left[ {2819.6 - 2724.84} \right]\\&= \frac{1}{{50}} \times 94.76\\&= 1.89\end{align}\]

Standard variation \({\sigma _2}\left( {weight} \right) = \sqrt {1.89} = 1.37\)

\[\begin{align}{\rm{C}}.{\rm{V}}\left( {{\text{weight}}} \right) &= \frac{{{\text{standard deviation}}}}{{mean}} \times 100\\&= \frac{{1.37}}{{5.22}} \times 100\\&= 26.24\end{align}\]

Thus, C.V of weights is greater than C.V of lengths.

Therefore, weights vary more than the lengths.

  
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