First slide

Introduction to statistics

Question

The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combined
sample of size 500 is

Moderate

A

64
B

65.2
C

67.2
D

64.2

Solution

Combined mean $\bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}}$
   $= \frac{200 \times 25 + 300 \times 10}{500} = 16$
Let $d_{1} = {\bar{x}}_{1} - \bar{x} = 25 - 16 = 9 \overset{}{}$ and
   $d_{2} = {\bar{x}}_{2} - \bar{x} = 10 - 16 = - 6$
Now we know that
   $\begin{array}{l} σ^{2} = \frac{n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}} \\ = \frac{200 (9 + 81) + 300 (16 + 36)}{500} = \frac{33600}{500} = 67.2 \end{array}$

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