Question

(c) on what percent of days does sally wait between 2.5 and 5.3 minutes for the bus?

Explanation:

Step1: Assume normal - distribution

Let's assume waiting - time is normally distributed with mean $\mu$ and standard deviation $\sigma$ (not given here). First, calculate $z$ - scores: $z_1=\frac{2.5 - \mu}{\sigma}$, $z_2=\frac{5.3 - \mu}{\sigma}$.

Step2: Use standard normal table

Find $P(z_1

Answer:

Without $\mu$ and $\sigma$, cannot give a numerical answer. Need to calculate $z$ - scores and use standard normal table.