Question

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1. Exercise 4.16 on page 73.
A beverage company makes weekly deliveries from its warehouse to a supermarket. The

delivery time varies based on interstate traffic and scheduling factors, but is normally distributed

with a mean of 17 hours and a standard deviation of 3 hours. What is the probability it will take

between 12 and 15 hours to make this weeks delivery?

2. Exercise 4.17 on page 73.

If the local restaurants in a small town record an average of 92 customers a day, with a

standard deviation of 11, how many customers does a restaurant need to serve to be in the top

25%?

3. Exercise 5.11 on page 86.

5.11. It is known that 30% of all customers of a major credit card pay their bills in full before any

interest charges are incurred. Suppose a simple random sample of 150 credit card holders is

selected. (Note that n

p= 150(0.3) = 45

>

5, and n(1-p) = 150(0.7) = 105

>

5, so the normal

approximation to the binomial is quite valid.)

a.   What is the probability that 30 or fewer customers pay their account balances in full before

any interest charges are incurred?

b.   What is the probability that between 40 and 50 customers pay their account balances in

full before any interest charges are incurred?

c.   What is the probability that more than 60 customers pay their account balances in full

before any interest charges are incurred?

Answer 1

1) Since μ=17 and σ=3.0 we have:
P ( 12<X<15 )=P ( 12−17< X−μ<15−17 )=P ((12−17)/3.0<(X−μ)/σ<(15−17)/3.0)
Since Z=(x−μ)/σ , (12−17)/3.0=−1.67 and (15−17)/3.0=−0.67 we have:
P ( 12<X<15 )=P ( −1.67<Z<−0.67 )
Use the standard normal table to conclude that:
P ( −1.67<Z<−0.67 )=0.2039

2) Z score corresponding to 75% is 0.67

Z=X-mean/sigma

Z*sigma= X-mean

X= 92+0.67*11

X= 92+7.37

X= 99.37

3) mean= np= 45 and

variance= npq= 31.5

std= sqrt(31.5)= 5.61

a) Since μ=45 and σ=31.5 we have:
P ( X<29.5 )=P ( X−μ<29.5−45 )=P ((X−μ)/σ<(29.5−45)/31.5)
Since (x−μ)/σ=Z and (29.5−45)/31.5=−0.49 we have:
P (X<29.5)=P (Z<−0.49)
Use the standard normal table to conclude that:
P (Z<−0.49)=0.3121

b) Since μ=45 and σ=31.5 we have:
P ( 40<X<50 )=P ( 40−45< X−μ<50−45 )=P ((40−45)/31.5<(X−μ)/σ<(50−45)/31.5)
Since Z=(x−μ)/σ , (40−45)/31.5=−0.16 and (50−45)/31.5=0.16 we have:
P ( 40<X<50 )=P ( −0.16<Z<0.16 )
Step 3: Use the standard normal table to conclude that:
P ( −0.16<Z<0.16 )=0.1272

c) Since μ=45 and σ=31.5 we have:
P ( X>60 )=P ( X−μ>60−45 )=P ((X−μ)/σ>(60−45)/31.5)
Since Z=(x−μ)/σ and (60−45)/31.5=0.48 we have:
P ( X>60 )=P ( Z>0.48 )
Use the standard normal table to conclude that:
P (Z>0.48)=0.3156