# 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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