Solution to HW problems
Ch.3
31. a. 607 is one standard deviation above the mean. Approximately 68% of the scores are between 407
and 607 with half of 68%, or 34%, of the scores between the mean of 507 and 607. Also, since the
distribution is symmetric, 50% of the scores are above the mean of 507. With 50% of the scores
above 507 and with 34% of the scores between 507 and 607, 50% - 34% = 16% of the scores are
above 607.
b.
707 is two standard deviations above the mean. Approximately 95% of the scores are between 307
and 707 with half of 95%, or 47.5%, of the scores between the mean of 507 and 707. Also, since the
distribution is symmetric, 50% of the scores are above the mean of 507. With 50% of the scores
above 507 and with 47.5% of the scores between 507 and 707, 50%- 47.5% = 2.5% of the scores are
above 707.
c.
Approximately 68% of the scores are between 407 and 607 with half of 68%, or 34%, of the scores
between 407 and the mean of 507.
d.
Approximately 95% of the scores are between 307 and 707 with half of 95%, or 47.5%, of the scores
between 307 and the mean of 507. Approximately 68% of the scores are between 407 and 607 with
half of 68%, or 34%, of the scores between the mean of 507 and 607. Thus, 47.5% + 34% = 81.5%
of the scores are between 307 and 607.
32. a.
z
x

x

2300  3100
 .67
1200

4900  3100
 1.50
1200
b.
z
c.
$2300 is .67 standard deviations below the mean. $4900 is 1.50 standard deviations above the mean.
Neither is an outlier.
d.
z

x


13000  3100
 8.25
1200
$13,000 is 8.25 standard deviations above the mean. This cost is an outlier.
34. a.
x
s
b.
z
xi 765

 76.5
n
10
( xi  x )2
442.5

7
n 1
10  1
x  x 84  76.5

 1.07
s
7
Approximately one standard deviation above the mean. Approximately 68% of the scores are within
one standard deviation. Thus, half of (100-68), or 16%, of the games should have a winning score of
84 or more points.
z
x  x 90  76.5

 1.93
s
7
Approximately two standard deviations above the mean. Approximately 95% of the scores are
within two standard deviations. Thus, half of (100-95), or 2.5%, of the games should have a winning
score of more than 90 points.
c.
x
s
xi 122

 12.2
n
10
( xi  x )2
559.6

 7.89
n 1
10  1
Largest margin 24: z 
35. a.
x
xi 79.86

 3.99
n
20
Median =
b.
x  x 24  12.2

 1.50 . No outliers.
s
7.89
4.17  4.20
 4.185 (average of 10th and 11th values)
2
Q1 = 4.00 (average of 5th and 6th values)
Q3 = 4.50 (average of 15th and 16th values)
( xi  x )2
12.51

 0.81
n 1
19
c.
s
d.
The distribution is significantly skewed to the left.
e.
Allison One: z 
4.12  3.99
 0.16
0.81
2.32  3.99
 2.06
0.81
The lowest rating is for the Bose 501 Series. Its z-score is:
Omni Audio SA 12.3: z 
f.
z
2.14  3.99
 2.28
0.81
This is not an outlier so there are no outliers.
46. a.
18
16
14
12
y
10
8
6
4
2
0
0
5
10
15
20
25
x
b.
Positive relationship
c/d. xi  80
x
80
 16
5
 ( xi  x )( yi  y )  106
yi  50
( xi  x )( yi  y ) 106

 26.5
n 1
5 1
sx 
( xi  x )2
272

 8.2462
n 1
5 1
sy 
( yi  y )2
86

 4.6368
n 1
5 1
sxy
sx s y

26.5
 0.693
(8.2462)(4.6368)
A positive linear relationship
Ch.4
P36 
3.
6!
 (6)(5)(4)  120
(6  3)!
BDF BFD DBF DFB FBD FDB
4.
a.
50
 10
5
 ( xi  x ) 2  272
sxy 
rxy 
y
 ( yi  y ) 2  86
30
1st Toss
2nd Toss
3rd Toss
H
T
H
(H,H,T)
T
H
H
T
H
T
T
H
T
H
T
b.
(H,H,H)
(H,T,H)
(H,T,T)
(T,H,H)
(T,H,T)
(T,T,H)
(T,T,T)
Let: H be head and T be tail
(H,H,H) (T,H,H)
(H,H,T) (T,H,T)
(H,T,H) (T,T,H)
(H,T,T) (T,T,T)
c.
The outcomes are equally likely, so the probability of each outcomes is 1/8.
27.
Big Ten
Pac-10
Yes
No
a.
P(Neither) =
b.
P(Either) =
c.
P(Both) =
29. a.
No
3645
6823
10,468
6823
 .51
13, 429
2961
4494
849


 .49
13, 429 13, 429 13, 429
849
 .06
13, 429
1033
 .36
P(E) =
2851
854
 .30
2851
964
 .34
P(D) =
2851
P(R) =
b.
Yes
849
2112
2,961
Yes; P(E  D) = 0
4494
8935
13,429
1033
 .43
2375
c.
Probability =
d.
964(.18) = 173.52
Rounding up we get 174 of deferred students admitted from regular admission pool.
Total admitted = 1033 + 174 = 1207
P(Admitted) = 1207/2851 = .42
Let C = event consumer uses a plastic card
B = event consumer is 18 to 24 years old
Bc = event consumer is over 24 years old
37.
Given information:
P(C)  .37
P(B C)  .19
P(Bc C)  .81
P(B)  .14
a.
P(C B) 
P(C  B)
P(B)
but P(C  B) is unknown. So first compute
P(C  B)  P(C) P(B C)
 .37(.19)  .0703
Then
P(C B) 
b.
P(C  B) .0703

 .5021
P(B)
.14
P(C Bc ) 
P(C  Bc )
P(Bc )
but P(C  Bc ) and P(Bc ) are unknown. However, they can be computed as follows.
P(C  Bc )  P(C)P(Bc C)
 .37(.81)  .2997
P(Bc )  1- P(B)  1  .14  .86
Then
P(C Bc ) 
P(C  Bc ) .2997

 .3485
.86
P(Bc )
c. There is a higher probability that the younger consumer, age 18 to 24, will use plastic when
making a purchase. The probability that the 18 to 24 year old consumer uses plastic is
.5021 and the probability that the older than 24 year old consumer uses plastic is .3485.
Note that there is greater than .50 probability that the 18 to 24 years old consumer will use
plastic.
d. Companies such as Visa, Mastercard and Discovery want their cards in the hands of
consumers who will have a high probability of using the card. So yes, these companies
should get their cards in the hands of young consumers even before these consumers have
established a credit history. The companies should place a low limit of the amount of credit
charges until the young consumer has demonstrated the responsibility to handle higher
credit limits.