Question

Let a continuous random variable $X$ follow Normal distribution with mean $\mu$ and variance $\sigma^2$. Let $Z = \dfrac{X - \mu}{\sigma}$. If $P(Z>Z_1) = 0.12$ and $P(Z>Z_2) = 0.76$, then $P(Z_2<Z<Z_1) =$

• A. 0.88
• B. 0.64
• C. 0.38
• D. 0.62

Hint:

Use the complement rule and properties of standard normal distribution: \(P(a<Z<b) = P(Z<b) - P(Z<a)\).

Answer 1

The Correct Option is B

We have a continuous random variable $X$ following a Normal distribution with mean $\mu$ and variance $\sigma^2$. The variable $Z = \dfrac{X - \mu}{\sigma}$ follows a standard normal distribution, i.e., $Z \sim N(0,1)$. We are given:

• $P(Z>Z_1) = 0.12$
• $P(Z>Z_2) = 0.76$

We need to find $P(Z_2<Z<Z_1)$.

Step 1: Find $P(Z<Z_1)$. Given $P(Z>Z_1) = 0.12$, using the standard normal property $P(Z<Z_1) = 1 - P(Z>Z_1) = 1 - 0.12 = 0.88$.

Step 2: Find $P(Z<Z_2)$. Given $P(Z>Z_2) = 0.76$, thus $P(Z<Z_2) = 1 - 0.76 = 0.24$.

Step 3: Calculate $P(Z_2<Z<Z_1)$. Use the result of the subtracted probabilities: $P(Z_2<Z<Z_1) = P(Z<Z_1) - P(Z<Z_2) = 0.88 - 0.24 = 0.64$.

Therefore, the probability $P(Z_2<Z<Z_1)$ is $\boxed{0.64}$.