Question

The number of values of α ∈ N such that the variance of 3, 7, 12, α, 43 – α. is a natural number is ______.

Options

• 0

• 2

• 5

• infinite

Answer 1

The number of values of α ∈ N such that the variance of 3, 7, 12, α, 43 – α. is a natural number is 0.

Explanation:

Given numbers: 3, 7, 12, α, 43 – α

Now, `barx = (3 + 7 + 12 + α + 43 - α)/5`

⇒ `barx` = 13

∵ Variance = `(sumx_1^2)/N - (barx)^2`

⇒ Variance = `(9 + 49 + 144 + α^2 + (43 - α)^2)/5`

⇒ Variance = `(202 + α^2 + α^2 + 1849 - 86α)/5 - 169`

⇒ Variance = `(2α^2 - 86α + 2051 - 845)/5`

⇒ Variance = `((2α^2 - α + 1) + (1205 - 85α))/5`

⇒ Variance = `((2α^2 - α + 1) + 5(241 - 17α))/5`

For variance to be a natural number `(2α^2 - α + 1)/5 ∈ N` 

⇒ 2α² – α + 1 – 5n = 0 must have solution as natural number

Now, discriminant of above quadratic equation is D = (–1)² – 4(2)(1 – 5n)

⇒ D = 40n – 7

So, D cannot be a perfect square as all perfect squares will be form of 4p or 4p + 1 for p ∈ N

∴ α can’t be natural number.