Arithmetic mean of two samples of volume 50 and 100 is 54 then find the composite mean of two sample
Answers
Answer:
Mean 51.6 (approx)
Std dev 7.3 (approx)
Step-by-step explanation:
The mean will just be the weighted average of the given means:
\text{mean} = \frac{50\times 54.1 + 100\times 50.3}{50+100} = \frac{7735}{150} \approx 51.6mean=
50+100
50×54.1+100×50.3
=
150
7735
≈51.6
For the standard deviation, this will depend upon whether you are using the so-called "sample standard deviation" here or the "population standard deviation". You're probably using the sample standard deviation, so I'll do that one first.
We're kinda doing weighted averages again, but we need to be careful of the fact that divisions (and so multiplications that "undo" previous divisions) are "degrees of freedom", not "number of values". So...
\sigma_{\text{sample}} = \sqrt{\frac{8^2\times(50-1) + 7^2\times(100-1)}{50+100-1}}= \sqrt{\frac{7987}{149}} \approx 7.32σ
sample
=
50+100−1
8
2
×(50−1)+7
2
×(100−1)
=
149
7987
≈7.32
Otherwise, in case you need it,
\sigma_{\text{population}} = \sqrt{\frac{8^2\times50 + 7^2\times100}{50+100}}= \sqrt{\frac{8100}{150}} = \sqrt{54} \approx 7.35σ
population
=
50+100
8
2
×50+7
2
×100
=
150
8100
=
54
≈7.35