For a random variable $X$ ($-\infty < X < \infty$) following normal distribution, the mean is $\mu = 100$. If the $P(X \geq 110)$, the find $P(90 \leq X \leq 110)$.

For a random variable $X$ ($-\infty < X < \infty$) following normal distribution, the mean is $\mu = 100$. If the $P(X \geq 110)$, the find $P(90 \leq X \leq 110)$. 

(A) $1 - 2\alpha$ 

(B) $1 - \alpha$ 

(C) $1 - \dfrac{\alpha}{2}$ 

(D) $2\alpha$ 

Answer: (A) $1 - 2\alpha$ 

Explanation: 

Here, $X$ is a random variable that follows a normal distribution with mean $\mu = 100$. 

Given, $P(X \geq 110) = \alpha$ 

Hence, $P(X \leq 90) = \alpha$ 

Therefore, $P(90 \leq X \leq 110) = 1 - 2\alpha$.