1. Given data

The 8 observations are: $4,5,6,6,7,8,x,y$ with mean $6$ and variance $\dfrac{9}{4}$.

We need to find $x^4+y^2$.

2. Use the mean

For 8 observations, mean is $\bar{x}=\frac{\text{sum of observations}}{8}=6$ So total sum is $8\times 6=48$

Now, $4+5+6+6+7+8+x+y=48$ The known sum is $4+5+6+6+7+8=36$ Hence, $36+x+y=48$ $x+y=12 \quad \text{...(1)}$

3. Use the variance

Variance is given by $\sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^2$

Given: $\sigma^2=\frac{9}{4},\quad \bar{x}=6,\quad n=8$ So, $\frac{\sum x_i^2}{8}-36=\frac{9}{4}$ $\frac{\sum x_i^2}{8}=36+\frac{9}{4}=\frac{144+9}{4}=\frac{153}{4}$ Thus, $\sum x_i^2=8\cdot \frac{153}{4}=2\cdot 153=306$

Now compute squares of known terms: $4^2+5^2+6^2+6^2+7^2+8^2=16+25+36+36+49+64=226$ Therefore, $226+x^2+y^2=306$ $x^2+y^2=80 \quad \text{...(2)}$

4. Find $xy$ using identity

We know: $(x+y)^2=x^2+y^2+2xy$ Using (1) and (2): $12^2=80+2xy$ $144=80+2xy$ $2xy=64$ $xy=32$

So $x$ and $y$ are roots of $t^2-(x+y)t+xy=0$ $t^2-12t+32=0$ Factorizing: $(t-4)(t-8)=0$ Hence, $x=4,\quad y=8$ (since $x<y$).

5. Compute the required value

$x^4+y^2=4^4+8^2=256+64=320$

6. Check options

• A: $162$
• B: $320$ ✅
• C: $674$
• D: $420$

Therefore, the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B.