Stat 50
Section 4.11 Solutions
Spring 2014
2. Let X be the thickness of a sheet of paper. Then   0.08 mm and   0.01 mm
a. Let S be the thickness of a 250 page book. Then S
N  20 , 0.025 and

20.2  20 
P  S  20.2   P  Z 
  P  Z  1.265   0.103
0.025 

b. P  Z  c   0.10  c  1.28 , so the 10th percentile for the thickness of a 250 page book is
20  1.28 0.025  19.80 mm .
c. No, because we haven’t been told that the thickness of individual pages, X, is normally distributed.
4. Let X be the tax paid.

1980  2000 
a. P  X  1980   P  Z 
  P  Z  1  0.8413
500 625 

b. Let Y be the number that paid a tax greater than $3000. Then Y ~ Bin  625, 0.1
P(Y  60 )  P( Z 
N  62.5, 56.25 , and
60.5  62.5
)  P( Z  0.27 )  0.6064 , using the continuity correction.
56.25
6. Let X be the warpage of a wafer. Then   1.3 mm and   0.1 mm

1.305  1.3 
P  X  1.305   P  Z 
  P  Z  0.707   0.2398
0
.
1
200


b. The 25th percentile of any normally distributed random variable is about 0.675 standard deviations
below the mean, so the 25th percentile for the mean warpage for samples of size 200 is
a.
 0.1 
1.3  0.675 
  1.295 mm
 200 
c. Solving
1.305  1.3
 1.645 for n yields n  1083
0.1
n
8. Let X be the amount of solution in a drum. Then   30.01 L and   0.1 L
a. Let S be the total amount of solution in 50 drums, in liters. Then, by the Central Limit Theorem
applied to totals, S N 1500.5 , 0.5  . Then
1500  1500.5 

P  S  1500   P  Z 
  P  Z  0.71  0.7611
0.5


b. Let T be the total amount of solution in 80 drums, in liters. Then, by the Central Limit Theorem
applied to totals, T N  2400.8 , 0.8  . Then
2401  2400.8 

P T  2401  P  Z 
  P  Z  0.22  0.5871
0.8


c. We want the 90th percentile of T. The 90th percentile of any normal distribution lies 1.28 standard
deviations above its mean, so the vat should contain 2400.8  1.28  0.8  2401.95 L
10. Let Y be the number that have a college degree. Then Y ~ Bin 100, 0.3
P(Y  35 )  P( Z 
N  30, 21 , and
35.5  30
)  P( Z  1.20 )  0.1151 , using the continuity correction.
21
14. Let X be the number of particles in 2 ml. Then X ~ Poisson( 60 ) ~ N ( 60 , 60 )
50  60
a. P( X  50 )  P( Z 
)  P( Z  1.29 )  0.9015 , where no continuity correction was used.
60
b. Let Y be the number of samples that contain more than 50 particles. Then Y ~ Bin 10, 0.9015 , and
P(Y  9 )  0.7419 .
c. Let Y be the number of samples that contain more than 50 particles. Then
Y ~ Bin 100, 0.9015 N  90.15, 8.88 , and
89.5  90.15
)  P( Z  0.22 )  0.5871 , using the continuity correction. (Note:
8.88
we assumed here that the normal approximation was appropriate although we didn't quite meet the
rule of 10 criterion specified by Navidi.)
P(Y  90 )  P( Z 
18. Let X and Y be the times for machines 1 and 2, respectively.
a. Let TX be the total time for machine 1. Then TX N  50,16  and

P (TX  55)  P  Z 

55  50 
  P Z  1.25   0.1056
16 
b. Let TY be the total time for machine 2. Then TY

P (TY  55)  P  Z 

N  60 , 25  and
55  60 
  P  Z  1.00   0.1587
25 
c. Let T TX TY be the total time for the two machines. Then T

P (T  115)  P  Z 

N  50  60,16  25   N 110, 41 , and
115  110 
  P  Z  0.78   0.2177 .
41 
d. Let D TX TY be the difference in total times for the two machines. Then

D N  50  60,16  25   N  10 ,41 , and P (D  0)  P  Z 

0  ( 10) 
  P  Z  1.56   0.0594
41 
20. Answers.
0.00012111000 
a.   15.3e
 4.0425
b. Let X be the number of emissions in a 25 minute interval. Then X ~ Poisson( 101.06 ) , with mean
101.06, and standard deviation 10.053
c. The mean and standard deviation of ̂ are 4.0425 and 0.4021, respectively.
d. ̂ =4.56
e. ̂ =3.60
f. P( 90  X  114 ) ~ P( 1.10  Z  1.29 )  0.7658 .


Note: In parts (d) and (e) I rounded ̂ to the nearest 0.04. Why?
Note: A continuity correction actually would have helped a bit here. According to my
calculator, the true Poisson probability is about 0.7856, whereas the continuity corrected
normal approximation produces a value of about 0.7843.