Subtopic: Poisson, normal and binomial distributions
Question Statement:
A continuous random variable $X$ follows a normal distribution with mean $\mu = 50$ and variance $\sigma^2 = 25$. Given that the cumulative standard normal distribution function $\Phi(1.645) = 0.95$, the value of $k$ such that $P(X > k) = 0.05$ is ________ (round off to two decimal places).
Answer Key:
58.23
Step-by-Step Solution & Explanation
Given $X \sim \mathcal{N}(\mu = 50, \sigma = \sqrt{25} = 5)$.
We are given $P(X > k) = 0.05 \implies P(X \le k) = 0.95$.
Standardizing $X$:
$P\left( Z \le \frac{k - 50}{5} \right) = 0.95$
From the standard normal table, $\Phi(1.645) = 0.95$.
Therefore:
$\frac{k - 50}{5} = 1.645 \implies k = 50 + 5(1.645) = 50 + 8.225 = 58.225 \approx 58.23$