Chapter 5 Solutions:
Section 5.1:
a.) x = number of days to fix defect
b.) Histogram
Number of Days to Fix Defect
30.00%
25.00%
Probability
1.)
20.00%
15.00%
10.00%
5.00%
0.00%
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18
x
c.) mean number of days to fix defect = m = 4.14415 days
d.) variance = s 2 » 8.3633 days2
e.) standard deviation = s » 2.892 days
f.) P ( x ³ 16 ) = 0.002 + 0.001+ 0.001 = 0.004
g.) It is unusual for a lens to take 16 days to have the defect fixed, since
P ( x ³ 16 ) < 0.05 .
h.) If it does take 16 days for a lens to be replaced, then there may be something
wrong with that lens, or the company may need to look at its processes to make
sure the processes are being done correctly.
2.)
3.)
–
x = winnings
Win
Lose
x
2500 -1 = $2499
-$1
P(x)
1
1
1- 0.0001 = 0.9999
=
= 0.0001
10 *10 *10 *10 10000
Expected value = 2499 *0.0001+ ( -1) *0.9999 = -$0.75
Section 5.2:
1.)
n = 14, p = 0.13, q = 1- 0.13 = 0.87
a.) P ( x = 5 ) = 14 C5 ( 0.13) ( 0.87 )
5
» 0.0212
14-5
b.) P ( x = 8 ) = 14 C8 ( 0.13) ( 0.87 )
8
» 1.062 ´10 -4
14-8
c.) P ( x = 12 ) = 14 C12 ( 0.13) ( 0.87 )
» 1.605 ´10 -9
d.)
P ( x £ 4 ) = P ( x = 0 ) + P ( x = 1) + P ( x = 2 ) + P ( x = 3) + P ( x = 4 )
12
14-12
= 14 C0 ( 0.13) ( 0.87 ) +
0
» 0.973
e.)
P ( x ³ 8) = P ( x = 8) +
14
+ 14 C4 ( 0.13) ( 0.87 )
4
10
+ P ( x = 14 )
= 14 C8 ( 0.13) ( 0.87 ) +
8
6
+ 14 C14 ( 0.13)
14
( 0.87 )0
» 1.18 ´10 -4
f.)
P ( x £ 12 ) = 1- P ( x > 12 )
= 1- P ( x ³ 13)
= 1- éë P ( x = 13) + P ( x = 14 ) ùû
= 1- éë 14 C13 ( 0.13)
13
( 0.87 )1 + 14 C14 ( 0.13)14 ( 0.87 )0 ùû
= 1- 3.73 ´ 10 -11
» 0.99999
2.)
3.)
n = 10 and p = 0.70
a.) P ( x = 2 ) = binompdf (10,.7,2 ) » 0.0014
b.) P ( x = 8 ) = binompdf (10,.7,8 ) » 0.2335
c.) P ( x = 7 ) = binompdf (10,.7, 7 ) » 0.2668
d.) P ( x £ 3) = binomcdf (10,.7, 3) » 0.0106
e.) P ( x ³ 7 ) = 1- P ( x < 7 ) = 1- P ( x £ 6 ) = 1- binomcdf (10,.7,6 ) » 0.6496
f.) P ( x £ 4 ) = binomcdf (10,.7, 4 ) » 0.0473
4.)
5.)
n = 17 and p = 0.63
a.) P ( x = 8 ) = binompdf (17,.63,8 ) » 0.0784
b.) P ( x = 15 ) = binompdf (17,.63,15 ) » 0.0182
c.) P ( x = 14 ) = binompdf (17,.63,14 ) » 0.0534
d.) P ( x £ 12 ) = binomcdf (17,.63,12 ) » 0.8142
e.) P ( x ³ 10 ) = 1- P ( x <10 ) = 1- P ( x £ 9 ) = 1- binomcdf (17,.63,9 ) » 0.7324
f.) P ( x £ 7 ) = binomcdf (17,.63, 7 ) » 0.0557
6.)
-
7.)
n = 15 and p = 0.10
a.) x = number of people who are left-handed
b.) There are a fixed number of trials, mainly 15 people. The trials are
independent, since on person’s handedness does not affect the next
persons. There are two outcomes, which are left-handed or not. The
probability of being left-handed is 10% for every trial.
c.) P ( x = 0 ) = binompdf (15,.1,0 ) » 0.2059
d.) P ( x = 7 ) = binompdf (15,.1, 7 ) » 2.770 ´10 -4
e.) P ( x ³ 2 ) = 1- P ( x < 2 ) = 1- P ( x £ 1) = 1- binomcdf (15,.1,1) » 0.4510
f.) P ( x £ 3) = binomcdf (15,.1,3) » 0.9444
g.)
P ( x ³ 7 ) = 1- P ( x < 7 )
= 1- P ( x £ 6 ) = 1- binomcdf (15,.1,6 ) » 3.106 ´10 -4
h.) It is unusual that 7 people would be left-handed, since the probability of at
least 7 people is less than 5%. It makes me think that there could be
something about the hand that someone uses to write with has something
to do with leadership if 7 of the last 15 U.S. Presidents were left-handed.
Or it could just be a coincidence.
8.)
9.)
n = 20 and p = 0.169
a.) x = number of defective lenses due to scratches
b.) There are a fixed number of trials, mainly 20 defective lenses. The trials
are independent since the reason one lens is affective doesn’t affect the
next lens. There are only two outcomes, defect due to scratches or not due
to scratches. The probability of a defective lens due to a scratch is the
same for each trial.
c.) P ( x = 0 ) = binompdf ( 20,.169,0 ) » 0.0247
d.) P ( x = 20 ) = binompdf ( 20,.169,20 ) » 3.612 ´10 -16
e.) P ( x ³ 3) = 1- P ( x < 3) = 1- P ( x £ 2 ) = 1- binomcdf ( 20,.169,2 ) » 0.6812
f.) P ( x £ 5 ) = binomcdf ( 20,.169,5 ) » 0.8926
g.)
P ( x ³ 10 ) = 1- P ( x < 10 )
= 1- P ( x £ 9 ) = 1- binomcdf ( 20,.169,9 ) » 6.711 ´10 -4
h.) It would be unusual for 10 defective lenses to be because of scratches,
since the probability is less than 5%. This may mean that there are more
defective lenses due to scratches than before. The company may want to
look at their processes.
10.) –
Section 5.3:
n = 6 and p = 0.13
a.) Probability distribution is found using binompdf(6,.13,x value)
x
p(x)
0
0.433626
1
0.388768
2
0.145230
3
0.028935
4
0.003243
5
0.000194
6
0.000005
b.) Histogram
Probabilty Distribution
p(x)
1.)
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
x
c.) This graph is skewed right.
d.) Mean = m = 6 *0.13 = 0.78
e.) Variance = s 2 = 6 *0.13*0.87 = 0.6786
f.) Standard deviation = s = 0.6786 » 0.8238
4
5
6
2.)
-
n = 7 and p = 0.50
a.) Probability distribution is found using binompdf(7,.50,x value)
x
0
1
2
3
4
5
6
7
p(x)
0.007813
0.054688
0.164063
0.273438
0.273438
0.164063
0.054688
0.007813
b.) Histogram
Probability Distribution
0.300000
0.250000
0.200000
p(x)
3.)
0.150000
0.100000
0.050000
0.000000
0
1
2
3
4
x
c.) This graph is symmetric.
d.) Mean = m = 7 *0.50 = 3.5
e.) Variance = s 2 = 7 *0.50 *0.50 = 1.75
f.) Standard deviation = s = 1.75 » 1.3229
5
6
7
n = 12 and p = 0.24
a.) x = number of patients who have been hospitalized for an acute myocardial
infarction who did not fill their cardiac medication by the seventh day of
being discharged
b.) Probability distribution is found using binompdf(12,.24,x value)
x
0
1
2
3
4
5
6
7
8
9
10
11
12
p(x)
0.037133
0.140716
0.244401
0.257264
0.182793
0.092358
0.034027
0.009210
0.001818
0.000255
0.000024
0.000001
0.000000
c.) Histogram
Probability Distribution
0.30
0.25
0.20
p(x)
4.)
5.)
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
x
d.) This graph is slightly skewed right.
e.) Mean = m = 12 *0.24 = 2.88
f.) Variance = s 2 = 12 *0.24 *0.76 = 2.1888
g.) Standard deviation = s = 2.1888 » 1.479
7
8
9
10
11
12
6.)
7.)
n = 52 and p = 0.14
a.) x = number of brown M&Ms in a milk chocolate packet
b.) Mean = m = 52 *0.14 = 7.28
c.) Variance = s 2 = 52 *0.14 *0.86 = 6.2608
d.) Standard deviation = s = 6.2608 » 2.502