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ASSUME THAT THE 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.1. IN A RANDOM SAMPLE OF 1000 CUSTOMERS, HOW MANY WAIT 18 MINUTES OR MORE BEFORE BEING SEATED.2. IN A RANDOM SAMPLE OF 500 CUSTOMERS, HOW MANY WAIT LESS THAN 9 MINUTES BEFORE BEING SEATED

Sagot :

Solution.

Calculate the z-score

The formula is shown below

[tex]\begin{gathered} \sigma=4 \\ \mu=16 \\ \end{gathered}[/tex][tex]\begin{gathered} Z_{18}=\frac{18-16}{4}=0.5 \\ P\left(x>0.5\right)=0.30854 \\ n=0.30854\text{ x 1000} \\ n=308.54 \\ n=309(nearest\text{ whole number\rparen} \end{gathered}[/tex]

Thus, 309 customers (to nearest whole number) wait 18 minutes or more before being seated

(ii)

[tex]\begin{gathered} Z_9=\frac{9-16}{4} \\ Z_9=-1.75 \\ P\left(x<-1.75\right)=0.040059 \\ n=0.040059\text{ x 500} \\ n=20.03 \\ n=20(nearest\text{ whole number\rparen} \end{gathered}[/tex]

Thus, 20 customers (to nearest whole number) wait less than 9 minutes before being seated

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