Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.2 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.2

Question 1.

A shooter missed his target in the last 10 trials by the following distance (mm) during the practice session:

20, 32, 24, 41, 18, 27, 15, 36, 35, 25

Find the quartile deviation and coefficient of quartile deviation of such distance missed by the shooter.

Answer:

Writing the measures of mistargets in ascending order :

15, 18, 20, 24, 25, 27, 32, 35, 36, 41

Here, n = 10

First Quartile :

Q_{1} = Value of \(\left(\frac{n+1}{4}\right)\)th observation

= Value of \(\left(\frac{10+1}{4}\right)\) = 2.75 th observation

= Value of 2nd observation +0.75 (Value of 3rd observation – Value of 2nd observation)

= 18 + 0.75 (20 – 18)

= 18 + 0.75 (2)

= 18 + 1.50 = 19.50 mm

Third Quartile :

Q_{3} = Value of 3\([latex]\left(\frac{n+1}{4}\right)\)[/latex] th observation

= Value of 3 (2.75) = 8.25th observation

= Value of 8th observation + 0.25 (Value of 9th observation – Value of 8th observation)

= 35 + 0.25 (36-35)

= 35 + 0.25 = 35.25 mm

Quartile deviation of measures of mistargets :

Q_{d} = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)

Putting Q_{3} = 35.25; Q1 = 19.50, we get

Q_{d} = \(\frac{35.25-19.50}{2}=\frac{15.75}{2}\) = 7.875 â‰ˆ 7.88

Coefficient of quartile deviation = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{\mathrm{Q}_{3}+\mathrm{Q}_{1}}\)

= \(\frac{35.25-19.50}{35.25+19.50}\)

= \(\frac{15.75}{54.75}\)

= 0.29

Question 2.

Find the quartile deviation and coefficient of quartile deviation of the marks from the following frequency distribution of marks of 43 students of a school;

Answer:

First Quartile :

Q_{1} = Value of \(\left(\frac{n+1}{4}\right)\)th observation

= Value of \(\left(\frac{43+1}{4}\right)=\frac{44}{4}\) = 11th observation

Referring to column cf,

Q_{1} = 20 marks

Third Quartile :

Q_{3} = Value of \(\left(\frac{n+1}{4}\right)\)th observation

= Value of 3(11)

= 33 rd observation Referring to column cf,

Q_{3} = 40 marks

Quartile deviation of marks:

Question 3.

The distribution of amount paid by 200 customers coming for snacks at a restaurant on a particular day is given below:

Find the quartile deviation and coefficient of quartile deviation of the amount paid by customers on the day.

Answer:

Amount paid â‚¹ | No. of customers f | Cumulative frequency cf |

0-50 | 25 | 25 |

50- 100 | 40 | 65 |

100-150 | 80 | 145 |

150-200 | 30 | 175 |

200 – 250 | 25 | 200 |

Total | n = 200 | – |

First Quartile:

Q_{1} Class = Class that Includes \(\left(\frac{n}{4}\right)\)th observation

= Class that includes \(\left(\frac{200}{4}\right)\) = 50th observation

Referring to column cf

Q_{1} class = 50 – 100

Now, Q_{1} = L + \(\frac{\left(\frac{n}{4}\right)-c f}{f}\) Ã— c

Putting L = 50: \(\left(\frac{n}{4}\right)\) = 50: cf = 25: f = 40 and c = 50 In the formula,

Third Quartile:

Q_{3} Class = Class that Includes 3\(\left(\frac{n}{4}\right)\)th observation

= Class that Includes 3 (50)th = 150th observation

Referring to column cf. Q_{3} class = 150 – 200

Now. Q_{3} = L + \(\frac{3\left(\frac{n}{4}\right)-c f}{f}\) Ã— c

Putting L = 150: 3\(\left(\frac{n}{4}\right)\) = 150: cf = 145; f = 30 and c = 50 in the formula.