Download Page 336: 12. a. You want in a distribution with mean and standard

Page 336: 12. a. You want P ( x  45, 000) in a distribution with mean   37, 764 and standard
deviation   5,100 . Calculate z 
x


45000  37764
 1.419. Then
5100
P( x  45, 000)  P( z  1.419)  0.0778
b. According to the central limit theorem sample means lie in a normal distribution with mean  x  
and standard deviation  x 

n
. You want P( x  38, 000) in a distribution with mean
 x  37, 764 and standard deviation  x 
x   x 38000  37764
5100
. Calculate z 

 0.401
5100
x
75
75
Then P( x  38, 000)  P( z  0.401) = normalcdf(0.401, 10000) = 0.3442
15. a. You want P ( x  670) in a distribution with mean   660 and standard deviation   35 .
x
670  660
 .2857

35
Then P( x  670)  P( z  0.2857)  normalcdf(0.2857,10000) = 0.3875
Calculate z 

b. According to the central limit theorem sample means lie in a normal distribution with mean  x  
and standard deviation  x 

n
. You want P ( x  670) in a distribution with mean  x  660 and
x   x 670  660
35
. Calculate z 

 0.9035
35
x
10
10
Then P( x  670)  P( z  0.904)  normalcdf(0.9035,10000) = 0.1831
c. The answer in part a is greater because the z - score in part a is less than the z - score in part b. The
higher a z - score is, the less likely it is to get a value with a higher z - score.
standard deviation  x 
20. a. You want P( x  50, 000) in a distribution with mean  x  51803 and standard deviation
x   x 50000  51803
4850
. Calculate z 

 −2.1677
4850
x
34
34
Then P( x  50000)  P( z  2.1667)  normalcdf(−2.1667,10000) = 0.9849
x 
b. a. You want P( x  48, 000) in a distribution with mean  x  51803 and standard deviation
x   x 48000  51803
4850
. Calculate z 

 −4.5722
4850
x
34
34
Then P( x  48000)  P( z  4.5722)  normalcdf(−10000, −4.5722) = 0.0000024
x 
22. a. You want P(120  x  121.8) in a distribution with mean   120 and standard deviation
  5.6 . Calculate z for 120 and 121.8
120: z  0 because 120 is the mean!
x
121.8  120
 0.3214

5.6
Then P(120  x  121.8)  P(0  z  0.3214)  normalcdf(0, 0.3214) = 0.1260
121.8: z 

b. According to the central limit theorem sample means lie in a normal distribution with mean  x  
and standard deviation  x 

n
. You want P(120  x  121.8) in a distribution with mean
 x  120 and standard deviation  x 
5.6
.
30
Calculate z for x = 120 and x = 121.8
120: z  0 because 120 is the mean!
x  x
121.8  120
 1.7605
5.6
x
30
Then P(120  x  121.8)  P(0  z  1.7605)  0.4608 normalcdf(0, 0.3214) = 0.1260
121.8: z 

c. The answer in part a is less because the z - score in part a is less than the z - score in part b. The
lower a z - score is, the less likely it is to get a value with a lower z - score.
24. I am not getting the book's answer in this one. Can anyone see something stupid? Books are wrong
more often than you think!
a. You want P(36  x  37.5) in a distribution with mean   36.2 and standard deviation   3.7 .
Calculate z for 36 and 37.5
x
36  36.2
 −0.0541

3.7
x   37.5  36.2

 0.3514
37.5: z 

3.7
Then P(36  x  37.5)  P(0.0541  z  0.3514)  normalcdf(−0.0541, 0.3514) = 0.1589
36: z 

b. According to the central limit theorem sample means lie in a normal distribution with mean  x  
and standard deviation  x 

. You want P (36  x  37.5) in a distribution with mean  x  36.2
n
3.7
and standard deviation  x 
.
15
Calculate z for x = 36 and x = 37.5
x  x
36  36.2
 −0.2094
3.7
x
15
x   x 37.5  36.2
x = 37.5: z 

 1.3608
3.7
x
15
Then P(36  x  37.5)  P(0.2094  z  1.3608)  normalcdf(−.2904, 1.3608) = 0.5275
x = 36: z 
