Given data

We have two groups of observations:

• Group I: size $n_1=10$, mean $\mu_1=2$, variance $\sigma_1^2=2$
• Group II: size $n_2=n$, mean $\mu_2=3$, variance $\sigma_2^2=1$

The variance of the combined data is given as $\sigma^2=\frac{17}{9}.$

Formula for combined mean

If the combined mean is $\mu$, then \mu=\frac{n_1\mu_1+n_2\mu_2}{n_1+n_2}=rac{10\cdot 2+n\cdot 3}{10+n}=rac{20+3n}{10+n}.

Formula for combined variance

The combined variance is $\sigma^2=\frac{n_1\big(\sigma_1^2+(\mu_1-\mu)^2\big)+n_2\big(\sigma_2^2+(\mu_2-\mu)^2\big)}{n_1+n_2}.$

Substitute the known values: $\frac{17}{9}=\frac{10\left(2+(2-\mu)^2\right)+n\left(1+(3-\mu)^2\right)}{10+n}.$

Compute $2-\mu$ and $3-\mu$

Since $\mu=\frac{20+3n}{10+n},$ we get 2-\mu=2-\frac{20+3n}{10+n}=rac{20+2n-20-3n}{10+n}=-\frac{n}{10+n}, so $(2-\mu)^2=\frac{n^2}{(10+n)^2}.$

Also, 3-\mu=3-\frac{20+3n}{10+n}=rac{30+3n-20-3n}{10+n}=\frac{10}{10+n}, so $(3-\mu)^2=\frac{100}{(10+n)^2}.$

Substitute into the variance formula

$\frac{17}{9}=\frac{10\left(2+\frac{n^2}{(10+n)^2}\right)+n\left(1+\frac{100}{(10+n)^2}\right)}{10+n}.$

Expand the numerator: $=\frac{20+\frac{10n^2}{(10+n)^2}+n+\frac{100n}{(10+n)^2}}{10+n}.$

Combine terms: $=\frac{20+n+\frac{10n^2+100n}{(10+n)^2}}{10+n}.$

Factor the fraction term: $10n^2+100n=10n(n+10),$ hence $\frac{10n^2+100n}{(10+n)^2}=\frac{10n(n+10)}{(10+n)^2}=\frac{10n}{10+n}.$

Therefore, $\frac{17}{9}=\frac{20+n+\frac{10n}{10+n}}{10+n}.$

Simplify

Write the numerator over a common denominator: $20+n+\frac{10n}{10+n}=\frac{(20+n)(10+n)+10n}{10+n}.$

So the whole expression becomes $\frac{17}{9}=\frac{(20+n)(10+n)+10n}{(10+n)^2}.$

Expand: $(20+n)(10+n)=200+30n+n^2,$ hence $\frac{17}{9}=\frac{n^2+40n+200}{(n+10)^2}.$

Solve for $n$

Cross-multiply: $9(n^2+40n+200)=17(n+10)^2.$

Expand both sides: $9n^2+360n+1800=17(n^2+20n+100)=17n^2+340n+1700.$

Rearranging: $0=17n^2+340n+1700-9n^2-360n-1800,$ $0=8n^2-20n-100.$

Divide by $4$: $2n^2-5n-25=0.$

Solve: $2n^2-5n-25=(2n+5)(n-5)=0.$

Thus, $n=5 \quad \text{or} \quad n=-\frac{5}{2}.$

Since size of a set cannot be negative, $n=5.$

Comparison with stored answer

The derived answer is $5$, which matches the stored correct answer.