solution - Millersville University
solution - Millersville University
solution - Millersville University
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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