The number of values of a in N such that the | vedclass

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(A) Given data: $3, 7, 12, a, 43-a$. Mean $\bar{x} = 13$.Sum of observations $= 3 + 7 + 12 + a + 43 - a = 65$.Mean $= \frac{65}{5} = 13$ (Verified).Va

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$0$

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(A) Given data: $3, 7, 12, a, 43-a$. Mean $\bar{x} = 13$.Sum of observations $= 3 + 7 + 12 + a + 43 - a = 65$.Mean $= \frac{65}{5} = 13$ (Verified).Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.$\sigma^2 = \frac{3^2 + 7^2 + 12^2 + a^2 + (43-a)^2}{5} - 13^2$.$\sigma^2 = \frac{9 + 49 + 144 + a^2 + 1849 - 86a + a^2}{5} - 169$.$\sigma^2 = \frac{2a^2 - 86a + 2051}{5} - 169 = \frac{2a^2 - 86a + 2051 - 845}{5} = \frac{2a^2 - 86a + 1206}{5}$.For $\sigma^2$ to be a natural number,$2a^2 - 86a + 1206$ must be divisible by $5$.$2a^2 - 86a + 1206 \equiv 0 \pmod{5} \Rightarrow 2a^2 - a + 1 \equiv 0 \pmod{5}$.Testing $a \pmod{5}$:If $a \equiv 0, 2(0)^2 - 0 + 1 = 1 
ot\equiv 0$.If $a \equiv 1, 2(1)^2 - 1 + 1 = 2 
ot\equiv 0$.If $a \equiv 2, 2(4) - 2 + 1 = 7 \equiv 2 
ot\equiv 0$.If $a \equiv 3, 2(9) - 3 + 1 = 16 \equiv 1 
ot\equiv 0$.If $a \equiv 4, 2(16) - 4 + 1 = 29 \equiv 4 
ot\equiv 0$.Since no value of $a$ satisfies the condition,the number of values is $0$.

The number of values of $a \in N$ such that the variance of $3, 7, 12, a, 43-a$ is a natural number is (Mean $= 13$).

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