1. Given distribution

The values are:

$x_i: 2,4,6,8,10,12,14,16$

with corresponding frequencies

$f_i: 4,4,\alpha,15,8,\beta,4,5$

Given:

• Mean $=9$
• Variance $=15.08$

We need to find:

$\alpha^2+\beta^2-\alpha\beta$

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2. Use the mean formula

Let total frequency be $N$.

$N=4+4+\alpha+15+8+\beta+4+5=40+\alpha+\beta$

Now,

$\sum f_i x_i = 4(2)+4(4)+\alpha(6)+15(8)+8(10)+\beta(12)+4(14)+5(16)$

Compute the constant part:

$4(2)=8,\quad 4(4)=16,\quad 15(8)=120,\quad 8(10)=80,\quad 4(14)=56,\quad 5(16)=80$

So,

$\sum f_i x_i = 8+16+6\alpha+120+80+12\beta+56+80$

$\sum f_i x_i = 360+6\alpha+12\beta$

Since mean is $9$,

$\frac{360+6\alpha+12\beta}{40+\alpha+\beta}=9$

So,

$360+6\alpha+12\beta = 360+9\alpha+9\beta$

$6\alpha+12\beta=9\alpha+9\beta$

$3\beta=3\alpha$

$\alpha=\beta$

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3. Use the variance formula

Variance is

$\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2$

Given mean $=9$ and variance $=15.08$,

$\frac{\sum f_i x_i^2}{N} = 15.08+9^2 = 15.08+81 = 96.08$

Now compute $\sum f_i x_i^2$:

$\sum f_i x_i^2 = 4(2^2)+4(4^2)+\alpha(6^2)+15(8^2)+8(10^2)+\beta(12^2)+4(14^2)+5(16^2)$

$=4(4)+4(16)+\alpha(36)+15(64)+8(100)+\beta(144)+4(196)+5(256)$

Compute constants:

$16+64+960+800+784+1280 = 3904$

Thus,

$\sum f_i x_i^2 = 3904+36\alpha+144\beta$

Since $\alpha=\beta$, let

$\alpha=\beta=t$

Then

$N=40+2t$

and

$\sum f_i x_i^2 = 3904+180t$

Using

$\frac{3904+180t}{40+2t}=96.08$

Multiply:

$3904+180t = 96.08(40+2t)$

$3904+180t = 3843.2+192.16t$

$60.8 = 12.16t$

$t=5$

Hence,

$\alpha=\beta=5$

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4. Required value

$\alpha^2+\beta^2-\alpha\beta = 5^2+5^2-5\cdot 5$

$=25+25-25=25$

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5. Comparison with stored answer

Stored correct answer: $25$

Our derived answer is also $25$, so they agree.