Question

at a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 30 minutes and a standard deviation of 2 minutes. what is the probability that a randomly selected customer will have to wait between 26 minutes and 32 minutes, to the nearest thousandth? statistics calculator

Explanation:

Step1: Identify the distribution parameters

The waiting time \( X \) is normally distributed with mean \( \mu = 30 \) minutes and standard deviation \( \sigma = 2 \) minutes, i.e., \( X \sim N(30, 2^2) \). We need to find \( P(26 < X < 32) \).

Step2: Standardize the values

We use the z - score formula \( z=\frac{x - \mu}{\sigma} \).
For \( x = 26 \):
\( z_1=\frac{26 - 30}{2}=\frac{- 4}{2}=-2 \)
For \( x = 32 \):
\( z_2=\frac{32 - 30}{2}=\frac{2}{2}=1 \)

So, \( P(26 < X < 32)=P(-2 < Z < 1) \), where \( Z \) is the standard normal variable.

Step3: Use the properties of the standard normal distribution

We know that \( P(-2 < Z < 1)=P(Z < 1)-P(Z < - 2) \).
From the standard normal table (or calculator):
\( P(Z < 1)\approx0.8413 \)
\( P(Z < - 2)\approx0.0228 \)

Step4: Calculate the probability

\( P(-2 < Z < 1)=0.8413 - 0.0228 = 0.8185\approx0.819 \) (to the nearest thousandth)

Answer:

\( 0.819 \)