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(b) In a random sample of 750 telephone users, what is the probability that more than 8%<br />

are “cell-phone only”?<br />

The Z-score associated with a proportion of 0.08 is<br />

Therefore<br />

z =<br />

0.08 − 0.07<br />

0.00931665<br />

≈ 1.07.<br />

P(ˆp > 0.08) = P(Z > 1.07)<br />

= 1 − P(Z < 1.07)<br />

= 1 − 0.8577<br />

= 0.1423.<br />

(c) Would it be unusual if a random sample of 750 adults results in 40 or fewer being<br />

“cell-phone only”?<br />

The sample proportion is<br />

ˆp = 40<br />

≈ 0.053.<br />

750<br />

The Z-score associated with this sample proportion is<br />

Therefore<br />

z =<br />

0.053 − 0.07<br />

0.00931665<br />

≈ −1.79.<br />

P(ˆp < 0.053) = P(Z < −1.79)<br />

= 0.0367.<br />

Since this probability is less than 5%, it would be unusual.<br />

Page 418, Exercise 30<br />

Billing Process<br />

For a certain billing process, the number of days for customers to pay their bill from<br />

the date of invoice is approximately normally distributed, with a mean of µ = 47 days and<br />

σ = 11 days. A random sample of 10 bills from the billing process during the month of June<br />

results in the following data:<br />

55 45 45 42 65<br />

58 35 36 34 60

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