Question:
Consider the statistics of two sets of observations as follows :
Size Mean Variance Observation I 10 2 2 Observation II n 3 1
If the variance of the combined set of these two observations is \(\frac{17}{9}\), then the value of n is equal to ___________.
Consider the statistics of two sets of observations as follows :
If the variance of the combined set of these two observations is \(\frac{17}{9}\), then the value of n is equal to ___________.
| Size | Mean | Variance | |
| Observation I | 10 | 2 | 2 |
| Observation II | n | 3 | 1 |
If the variance of the combined set of these two observations is \(\frac{17}{9}\), then the value of n is equal to ___________.
Show Hint
Use the combined-variance formula \(\sigma^2=\frac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2}+\frac{n_1n_2(\bar x_1-\bar x_2)^2}{(n_1+n_2)^2}\).
Updated On: Nov 4, 2026
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Correct Answer: 5
Solution and Explanation
Combined size \(N=10+n\).
\(\sigma^2=\frac{10\cdot2+n\cdot1}{10+n}+\frac{10n(2-3)^2}{(10+n)^2}=\frac{20+n}{10+n}+\frac{10n}{(10+n)^2}=\frac{17}{9}\).
Multiply by \(9(10+n)^2\): \(9(20+n)(10+n)+90n=17(10+n)^2\).
\(1800+270n+9n^2+90n=1700+340n+17n^2\Rightarrow8n^2-20n-100=0\Rightarrow2n^2-5n-25=0\).
\((2n+5)(n-5)=0\Rightarrow n=5\) (n positive).
Answer: 5
\(\sigma^2=\frac{10\cdot2+n\cdot1}{10+n}+\frac{10n(2-3)^2}{(10+n)^2}=\frac{20+n}{10+n}+\frac{10n}{(10+n)^2}=\frac{17}{9}\).
Multiply by \(9(10+n)^2\): \(9(20+n)(10+n)+90n=17(10+n)^2\).
\(1800+270n+9n^2+90n=1700+340n+17n^2\Rightarrow8n^2-20n-100=0\Rightarrow2n^2-5n-25=0\).
\((2n+5)(n-5)=0\Rightarrow n=5\) (n positive).
Answer: 5
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