The mean and variance of the data 4, 5, 6, 6, | vedclass

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(B) Given the mean $\bar{x} = 6$ for $8$ observations:$\frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$$36 + x + y = 48 \Rightarrow x + y = 12$  $(1)$Give

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$320$

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(B) Given the mean $\bar{x} = 6$ for $8$ observations:$\frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$$36 + x + y = 48 \Rightarrow x + y = 12$  $(1)$Given the variance $\sigma^{2} = \frac{9}{4}$:$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} = \frac{9}{4}$$\frac{4^{2} + 5^{2} + 6^{2} + 6^{2} + 7^{2} + 8^{2} + x^{2} + y^{2}}{8} - 6^{2} = \frac{9}{4}$$\frac{16 + 25 + 36 + 36 + 49 + 64 + x^{2} + y^{2}}{8} - 36 = \frac{9}{4}$$\frac{226 + x^{2} + y^{2}}{8} = 36 + 2.25 = 38.25$$226 + x^{2} + y^{2} = 306 \Rightarrow x^{2} + y^{2} = 80$  $(2)$From $(1)$,$y = 12 - x$. Substituting into $(2)$:$x^{2} + (12 - x)^{2} = 80$$x^{2} + 144 - 24x + x^{2} = 80$$2x^{2} - 24x + 64 = 0 \Rightarrow x^{2} - 12x + 32 = 0$$(x - 4)(x - 8) = 0$Since $x Calculating $x^{4} + y^{2}$:$4^{4} + 8^{2} = 256 + 64 = 320$.

Daily wage (Rs.)	$30$-$40$	$40$-$50$	$50$-$60$	$60$-$70$	$70$-$80$	$80$-$90$
No. of workers	$17$	$28$	$21$	$15$	$13$	$6$

Class:	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq:	$\alpha$	$110$	$54$	$30$	$\beta$

The mean and variance of the data $4, 5, 6, 6, 7, 8, x, y$ where $x < y$ are $6$ and $\frac{9}{4}$ respectively. Then $x^{4} + y^{2}$ is equal to

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