Subtopic: Probability distributions – binomial, Poisson, normal, exponential and Weibull distributions
Question Statement:
Referring to the standard normal distribution curve $f(z)$ shown in the accompanying diagram with mean $\mu = 0$ and standard deviation $\sigma = 1$, the cumulative probability percentage within the interval $[-1.96, +1.96]$ (representing the $95\%$ confidence interval) is ________ $%$ (round off to one decimal place).
Answer Key:
95.0
Step-by-Step Solution & Explanation
For a standard normal random variable $Z \sim N(0, 1)$, the cumulative distribution function is denoted by $\Phi(z)$.
1. From standard normal probability tables:
$\Phi(1.96) \approx 0.9750$
$\Phi(-1.96) = 1 - \Phi(1.96) = 1 - 0.9750 = 0.0250$
2. The probability enclosed in the interval $[-1.96, +1.96]$ is:
$P(-1.96 \le Z \le 1.96) = \Phi(1.96) - \Phi(-1.96) = 0.9750 - 0.0250 = 0.9500$
Expressed as a percentage: $95.0\%$.