For a group of 50 male workers, the mean and the standard deviation of their daily wages are Rs. 630 and Rs. 90 respectively and for a group of 40 female workers these are Rs. 540 and Rs. 60 respectively. Then, the

Correct answer: D. 90 n _ 1 = 50, ¯ x 1 = 630, σ _ 1 = 90 n _ 2 = 40, ¯ x 2 = 540, σ _ 2 = 60 ¯ x = (50 × 630 + 40 × 540)/(50 + 40) = (3150 + 2160)/(9) = (5310)/(9) = 590 Now, d _ 1 = ¯ x 1 - ¯ x , d _ 2 = ¯ x 2 - ¯ x d _ 1 = 630 - 590 = 40 d _ 2 = 540 - 590 = - 50 S.D. of the 90 workers = n _ 1 ( σ _ 1 ^ 2 + d _ 1 ^ 2 ) + n _ 2 ( σ _ 2 ^ 2 + d _ 2 ^ 2 ) n _ 1 + n _ 2 = 50 ( 8100 + 1600 ) + 40 ( 3600 + 2500 ) 90 = 5 ( 9700 ) + 4 ( 6100 ) 9 = 48500 + 24400 9 = (72900)/(9) = 8100 = 90

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Statistics

NTA Abhyas 2020 · Mathematics · Statistics

For a group of 50 male workers, the mean and the standard deviation...

For a group of

50

male workers, the mean and the standard deviation of their daily wages are

\text{Rs}. 630

and

\text{Rs}. 90

respectively and for a group of

40

female workers these are

\text{Rs}. 540

and

\text{Rs}. 60

respectively. Then, the standard deviation of all these

90

workers is

Options

A $60$
B $70$
C $80$
D $90$

Correct answer

D. $90$

Step-by-step solution

{n}_{1} = 50, \overset{¯}{x}1 = 630, {σ}_{1} = 90

{n}_{2} = 40, \overset{¯}{x}2 = 540, {σ}_{2} = 60

\overset{¯}{x} = \dfrac{50 \times 630 + 40 \times 540}{50 + 40} = \dfrac{3150 + 2160}{9}

= \dfrac{5310}{9} = 590

Now,

{d}_{1} = \overset{¯}{x}1 - \overset{¯}{x}, {d}_{2} = \overset{¯}{x}2 - \overset{¯}{x}

{d}_{1} = 630 - 590 = 40

{d}_{2} = 540 - 590 = - 50

S.D. of the

90

workers

=\sqrt{\dfrac{{n}_{1} ( {\sigma}_{1}^{2} + {d}_{1}^{2} ) + {n}_{2} ( {\sigma}_{2}^{2} + {d}_{2}^{2} )}{{n}_{1} + {n}_{2}}}

=\dfrac{\sqrt{50 ( 8100 + 1600 ) + 40 ( 3600 + 2500 )}}{90}

=\dfrac{\sqrt{5 ( 9700 ) + 4 ( 6100 )}}{9}

=\dfrac{\sqrt{48500 + 24400}}{9}

=\sqrt{\dfrac{72900}{9}} = \sqrt{8100} = 90

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