Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.3 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.3

Question 1.

The measurements of height (in centimeters) of 10 soldiers are given below:

Find the mean deviation of the heights of the soldiers.

Answer:

Here, n = 10

Height (in cm) | | x – x̄ | x̄ = 167.4 |

160 | 7.4 |

175 | 7.6 |

158 | 9.4 |

165 | 2.4 |

170 | 2.6 |

166 | 1.4 |

173 | 5.6 |

176 | 8.6 |

163 | 4.4 |

168 | 0.6 |

Σx = 1674 | Σ |x – x̄| = 50.0 |

Question 2.

The distribution of number of ball bearings used in machines of a factory is given below. Calculate the mean deviation and coefficient of mean deviation of number of ball bearings per machine.

Answer:

Mean:

x̄ = \(\frac{\Sigma f x}{n}=\frac{160}{20}\) = 8 ball bearing

Mean deviation of numbers of ball bearing:

MD = \(\frac{\Sigma f|x-\bar{x}|}{n}\)

= \(\frac{56}{20}\) = 28 ball bearing

Coefficient Mean deviation of ball bearing:

Coefficient of mean deviation = \(\frac{\mathrm{MD}}{\bar{x}}\)

= \(\frac{2.8}{8}\) = 0.35

Question 3.

Find the mean deviation and the coefficient of mean deviation of the distribution of talk time (in minutes) per call.

Answer:

Coefficient of Mean deviation of talk time : Coefficient of Mean deviation = \(\frac{\mathrm{MD}}{\bar{x}}=\frac{3.33}{7.3}\) = 0.46

Question 4.

Find the mean deviation and coefficient of mean deviation of number of TV sets using the following frequency distribution of TV sets sold in last 16 months in a town:

Answer:

Mean:

x̄ = \(\frac{\Sigma f x}{n}=\frac{960}{16}\) = 60 TV sets

Mean deviation of monthly sales of TV sets:

MD = \(\frac{\Sigma f|x-\bar{x}|}{n}=\frac{240}{16}\) = 15 TV sets

Coefficient of Mean deviation of monthly sales of TV sets:

Coefficient of mean deviation = \(\frac{M D}{\bar{x}}=\frac{15}{60}\) = 0.25

Question 5.

There are 50 boxes containing different number of units of an item in a factory. Find the mean deviation of number of units per box using the following distribution of the units:

Answer:

Mean:

x̄ = \(\frac{\Sigma f x}{n}=\frac{1670}{50}\) = 33.4 units

Mean deviation of number of units per box:

MD = \(\frac{\Sigma f|x-\bar{x}|}{n}=\frac{659.2}{50}\) = 13.18 units