Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.6 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.6

Question 1.

The information regarding marks of the students of two classes of a school is given below. Find the combined standard deviation of the marks obtained by the students.

Division A | Division B | |

No. of Students | 50 | 60 |

Mean marks | 60 | 48 |

Standard deviation | 10 | 12 |

Answer:

Class A:

n_{1} = No. of students = 50

x̄_{1} = Average marks = 60

s_{1} = Standard deviation = 10

Class B:

n_{2} = No. of students = 60

x̄_{2} = Average marks = 48

s_{2} = Standard deviation = 12

Combined mean of marks of students of Class A and Class B:

x̄_{c} = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

= \(\frac{(50 \times 60)+(60 \times 48)}{50+60}\)

= \(\frac{3000+2880}{110}=\frac{5880}{110}\) = 53.45 marks

Combined standard deviation of marks of students of Class A and Class B:

d_{1} = x̄_{1} – x̄_{c} = 60 – 53.45 = 6.55

∴ d_{1}^{2} = 42.9025

d_{2} = x̄_{2} – x̄_{c} = 48 – 53.45 = 5.45

∴d_{2}^{1} = 29.7025

s_{1} = 10 ∴ s_{1}^{2} = 100: s_{2} = 12 ∴ s_{2}^{2} = 144

Question 2.

The following information is available for two sections of a factory. Obtain the combined standard deviation of the production time.

Section A | Section B | |

No. of workers | 10 | 40 |

Mean production time per unit (minute) | 25 | 20 |

Variance | 16 | 25 |

Answer:

Section A:

n_{1} = No. of workers =10

x̄_{1} = Average production time per unit = 25

s_{1}^{1} = Variance =16

Section B :

n_{2} = No. of workers = 40

x̄_{2} = Average production time per unit = 20

s_{2}^{2} = Variance = 25

Combine mean of production time per unit of Section A and Section B:

x̄_{c} = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

= \(\frac{(10 \times 25)+(40 \times 20)}{10+40}\)

= \(\frac{250+800}{50}=\frac{1050}{50}\) = 21 minutes

Combined standard deviation of production time per unit of Section A and Section B :

d_{1} = x̄_{1} – x̄_{c} = 25 – 21 = 4

∴ d_{1}^{2} = 16

d_{2} = x̄_{2} – x̄_{c} = 20 – 21 = -1

∴d_{2}^{1} = 1

s_{1} = 4 ∴ s_{1}^{2} = 16: s_{2} = 5 ∴ s_{2}^{2} = 25