Question

WAITING TIMES FOR CUSTOMERS AT A POPULAR RESTAURANT BEFORE BEING SEATED ARE NORMALLY DISTRIBUTED WITH A MEAN OF 16 MINUTES AND STANDARD DEVIAITON OF 4 MINUTES.A) FIND THE PERCENT OF CUSTOMERS WHO WAIT FOR AT LEAST 14 MINUTES BEFORE BEING SEATED.B) FIND THE PERCENT OF CUSTOMERS WHO WAIT BETWEEN 12 AND 24 MINUTES BEFORE BEING SEATED.C) FIND THE PERCENT OF CUSTOMERS WHO WAIT AT LEAST 21 MINUTES BEFORE BEING SEATED

Answer 1

Solution.

Calculate the standard deviation

The formula is given below

`\begin{gathered} \sigma=4 \\ \mu=16 \\ \end{gathered}`
`(A)\text{ }Z_(14)=(14-16)/(4)=-0.5`

P(x>-0.5) = 0.69146

In percentage, P(x>Z) = 69.15% (2 decimal places)

THE PERCENT OF CUSTOMERS WHO WAIT FOR AT LEAST 14 MINUTES

BEFORE BEING SEATED is 69.15%

(B)

`\begin{gathered} Z_(12)=(12-16)/(4)=-1 \\ \end{gathered}`
`Z_(24)=(24-16)/(4)=2`

`\begin{gathered} P\left(x>1.25\right)=0.10565 \\ In\text{ percentage, we have 10.57\% \lparen2 decimal places\rparen} \end{gathered}`

Thus, THE PERCENT OF CUSTOMERS WHO WAIT AT LEAST 21 MINUTES BEFORE BEING SEATED is 10.57%