Question #169754

7.The daily delivery of mail at a large city firm follows a time pattern conforming to the Normal

Distribution, with a mean time of arrival of 9.40 am and with a standard deviation of 20 minutes.

Estimate the number of occasions during the 250 working days in the year when the mail arrives

a. before the main gates open at 9 am

b. after the arrival of the office staff at 9.20 am

c. between 10 am and 10.20 am


1
Expert's answer
2021-03-12T06:47:08-0500

Let X=X= difference in time of arrival with respect to 9.40 am in minutes.

XN(μ,σ2).X\sim N(\mu, \sigma^2). Then Z=XμσN(0,1)Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)

Given μ=0,σ=20 min\mu=0, \sigma=20\ min

a.

P(X<40)=P(Z<40020)P(X<-40)=P(Z<\dfrac{-40-0}{20})

=P(Z<2)0.02275=P(Z<-2)\approx0.02275

0.02275(250)=6(occasions)0.02275(250)=6(occasions)


b.

P(X>20)=1P(X20)P(X>-20)=1-P(X\leq -20)

=1P(Z20020)=1P(Z1)=1-P(Z\leq\dfrac{-20-0}{20})=1-P(Z\leq-1)0.84134\approx0.84134

0.84134(250)=210(occasions)0.84134(250)=210(occasions)


c.

P(20<X<40)=P(X<40)P(X20)P(20<X<40)=P(X<40)-P(X\leq 20)

=P(Z<40020)P(Z20020)=P(Z<\dfrac{40-0}{20})-P(Z\leq\dfrac{20-0}{20})




=P(Z<2)P(Z1)=P(Z<2)-P(Z\leq1)




0.977250.841340.13591\approx0.97725-0.84134\approx0.13591

0.13591(250)=34(occasions)0.13591(250)=34(occasions)



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