1. Use the condition $\sum f_i=62$

Given frequencies:

$$\begin{aligned} f_0&=k+2,\\ f_1&=2k,\\ f_2&=k^2-1,\\ f_3&=k^2-1,\\ f_4&=k^2+1,\\ f_5&=k-3. \end{aligned}$$

So,

$$(k+2)+2k+(k^2-1)+(k^2-1)+(k^2+1)+(k-3)=62.$$

Simplifying,

$$3k^2+4k-2=62$$ $$3k^2+4k-64=0.$$

Factorizing,

$$3k^2+16k-12k-64=0$$ $$(3k+16)(k-4)=0.$$

Hence,

$$k=4 \quad \text{or} \quad k=-\frac{16}{3}.$$

Since frequencies must be non-negative, $k=-\frac{16}{3}$ is invalid. Therefore,

$$k=4.$$

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2. Write the frequency table

Substitute $k=4$:

$$\begin{array}{c|cccccc} x_i & 0 & 1 & 2 & 3 & 4 & 5 \\\hline f_i & 6 & 8 & 15 & 15 & 17 & 1 \end{array}$$

Check:

$$6+8+15+15+17+1=62.$$

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3. Compute the mean $\mu$

$$\mu=\frac{\sum f_i x_i}{\sum f_i}.$$

Now,

$$\sum f_i x_i=0\cdot 6+1\cdot 8+2\cdot 15+3\cdot 15+4\cdot 17+5\cdot 1.$$ $$=0+8+30+45+68+5=156.$$

Thus,

$$\mu=\frac{156}{62}=\frac{78}{31}.$$

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4. Compute $\mu^2+\sigma^2$

Recall:

$$\sigma^2=E(X^2)-\mu^2.$$

So,

$$\mu^2+\sigma^2=E(X^2)=\frac{\sum f_i x_i^2}{\sum f_i}.$$

Hence it is easier to calculate $E(X^2)$ directly.

$$\sum f_i x_i^2=0^2\cdot 6+1^2\cdot 8+2^2\cdot 15+3^2\cdot 15+4^2\cdot 17+5^2\cdot 1.$$ $$=0+8+60+135+272+25=500.$$

Therefore,

$$\mu^2+\sigma^2=\frac{500}{62}=\frac{250}{31}\approx 8.0645.$$

So,

$$\left[\mu^2+\sigma^2\right]=\left[\frac{250}{31}\right]=8.$$

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5. Evaluate options

• A: $9$ ❌
• B: $8$ ✅
• C: $6$ ❌
• D: $7$ ❌

Therefore, the correct option is B.