Question

question at a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 46 minutes and a standard deviation of 5 minutes. what is the probability that a randomly selected customer will have to wait between 38 minutes and 55 minutes, to the nearest thousandth? statistics calculator answer attempt 1 out of 2 submit answer watch video show examples

Explanation:

Step1: Identify the distribution and parameters

The waiting time \( X \) is normally distributed with mean \( \mu = 46 \) minutes and standard deviation \( \sigma = 5 \) minutes. We need to find \( P(38 < X < 55) \).

Step2: Calculate the z - scores

The formula for the z - score is \( z=\frac{x-\mu}{\sigma} \).

For \( x = 38 \):
\( z_1=\frac{38 - 46}{5}=\frac{- 8}{5}=-1.6 \)

For \( x = 55 \):
\( z_2=\frac{55 - 46}{5}=\frac{9}{5} = 1.8 \)

Step3: Find the probabilities using z - table

We know that \( P(38 < X < 55)=P(-1.6 < Z < 1.8) \), where \( Z \) is the standard normal variable.

\( P(-1.6 < Z < 1.8)=P(Z < 1.8)-P(Z < - 1.6) \)

From the standard normal table, \( P(Z < 1.8)=0.9641 \) and \( P(Z < - 1.6)=0.0548 \)

Step4: Calculate the final probability

\( P(-1.6 < Z < 1.8)=0.9641 - 0.0548=0.9093 \)

Answer:

\( 0.909 \) (rounded to the nearest thousandth)