# ch05 - OoCities

```5.23
Let x be a normally distributed random variable having mean  = 30 and standard deviation
  5 . Find the z value for each of the following observed values of x:
In each case, explain what the s value tells us about how the observed value of x compares to the
mean
x 25 15 30 40 50
x
Z
z -1 -3 0 2 4

5.29
Weekly demand at a grocery store for a brand of breakfast cereal is normally distributed with a
mean of 800 boxes and a standard deviation of 75 boxes,
a. What is the .probability that weekly demand is
(1) 959 boxes or less?
(2) More than 1,004 boxes?
(3) Less than 650 boxes or greater than 950 boxes?
x
&lt;959 &gt;1004 &lt;650 and &gt;950
※ Z
z 2.12
2.72
2

p 0.9830 0.0033
0.0456
b. The store orders cereal from a distributor weekly. How many boxes should the store order
for a week to have only a 2.5 percent chance of running short of this brand of cereal durin&quot;
the week?
x  800
 947
※ z 0.025  1.96 
75
5.35
Recall from Exercise 5.32 that yearly returns on common stocks are normally distributed with a
mean of 12.4 percent and a standard deviation of 20.6 percent.
a. What percentage of yearly returns are at or below the 10th percentile of the distribution of
yearly returns? What percentage are at or above the 10th percentile? Find the 10th percentile of
the distribution of yearly returns.
※ Below 10 percentile = 10% and above 10 percentile=90%
x  12.4
 1.282, x  14.009%
※ Q10  z 0.1 
20.6
b. Find the first quartile. Q, and the third quartile, O,, of the distribution of yearly returns.
x  12.4
 0.6745, x  1.4947%
※ Q25  z 0.25 
20.6
x  12.4
 0.6745, x  26.2947%
※ Q75  z 0.75 
20.6
5.43
Under what condition may we use the normal approximation to the binomial distribution?
if np  5 and n(1  p )  5 , binominal random variable x 趨近於常態分配
5.47
An advertising agency conducted an ad campaign aimed at making consumers in an Eastern state
aware of a new product. Upon completion of the campaign, the agency claimed that 20 percent of
consumers in the state had become aware of the product. The product's distributor surveyed 1,000
consumers in the state and found that 150 were aware of the product.
a. Assuming that the ad agency's claim is true:
(1) Verify that we may use the normal approximation to the binomial.
np  5 and n(1  p )  5
1000  0.2  200and1000  08  800
(2) Calculate the mean.  , and the standard deviation,  , we should use in the normal
approximation.
  np  1000  0.2  200 and   npq  1000  0.2  0.8  12.65
(3) Find the probability that 150 or fewer consumers in a random sample of 1,000
consumers would be aware of the product.

p  p0
0.15  0.20
 3.95 p(z&lt;-3.95)=0.000
0.2  0.8  1000
p0 (1  p0 ) / n
b. Should the distributor believe the ad agency's claim? Explain.
NO, p&lt;0.001
z
=
5．57
The length of a particular telemarketing phone call, x has an exponential distribution with mean
equal to 1.5 minutes.
a.Write the formula for the exponential probability curve of x.
f ( x)  ex ( x  0) or f ( x)  0( x  other )
b. Sketch the probability curve of x .
C. Find the probability that the length of a randomly selected call will be
(1) No more than three minutes.
(2) Between one and two minutes.
(3) More than four minutes.
(4) Less than 30 seconds.
1
2
&lt;3
1&lt;x&lt;2
&gt;4
&lt;0.5



2
/
3

4
/
3

2

8
/
3
1.5 3
f (x ) 1  e
e
e
e
1  e 1 / 3
```