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STATISTICS FOR SOCIAL SCIENCESBUS 152
___________________________________________________________________________
SOME CONTINUOUS PROBABILITY DISTRIBUTIONSSOLUTIONS
Section 6.1
Problem 1 (6.1)
SOLUTIONS:a) P(Z< 1.57) = 0.9418b) P(Z> 1.84) = 1 0.9671 = 0.0329c) P(1.57 < Z< 1.84) = 0.9671 0.9418 = 0.0253
d) P(Z< 1.57) + P(Z> 1.84) = 0.9418 + (1 0.9671) = 0.9747_________________________________________________________________________________
Problem 2 (6.3)
SOLUTIONS:
a) P(Z< 1.08) = 0.8599b) P(Z> 0.21) = 1.0 0.4168 = 0.5832c) P(Z< 0.21) + P(Z> 0) = 0.4168 + 0.5 = 0.9168d) P(Z< 0.21) + P(Z> 1.08) = 0.4168 + (1 0.8599) = 0.5569
_________________________________________________________________________________
Problem 3 (6.5)
SOLUTIONS:
a) Z=X
=
75 100
10= 2.50
P(X> 75) = P(Z> 2.50) = 1 P(Z< 2.50) =1 0.0062 = 0.9938
b) Z=X
=
70 100
10= 3.00
P(X< 70) = P(Z< 3.00) = 0.00135
c) Z= X
= 8 0 1 0 01 0
= 2.00 Z= X
= 1 1 0 1 0 01 0
= 1.00
P(X< 80) = P(Z< 2.00) = 0.0228P(X> 110) = P(Z> 1.00) = 1 P(Z< 1.00) = 1.0 0.8413 = 0.1587P(X< 80) + P(X> 110) = 0.0228 + 0.1587 = 0.1815
d) P(Xlower < X< Xupper) = 0.80P( 1.28 < Z) = 0.10 and P(Z< 1.28) = 0.90
Z= 1.28 =Xlower100
1 0Z= +1.28 =
Xupper100
1 0
Xlower = 100 1.28(10) = 87.20 and Xupper = 100 + 1.28(10) = 112.80_________________________________________________________________________________
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Problem 4 (6.7)
SOLUTIONS:a) P(X< 25) = P(Z< 1.116) = 0.1322b) P(X> 50) = P(Z> 1.384) = 0.0832
c) P(30 < X< 40) = P(0.616 < Z< 0.384) = 0.3806
d) P(X< A) = 0.99 Z= 2.3263 =36.16
10
A A = $59.42
Note: The above answers are obtained using Excel. They may be slightly different whenTable E.2 is used.
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Problem 5 (6.11)
SOLUTIONS:
a) P(X< 180) = P(Z< 1.50) = 0.0668b) P(180 < X< 300) = P( 1.50 < Z< 1.50) = 0.9332 0.0668 = 0.8664c) P(110 < X< 180) = P( 3.25 < Z< 1.50) = 0.0668 0.00058 = 0.06622d) P(X< A) = 0.01 P(Z< 2.33) = 0.01
A = 240 2.33(40) = 146.80 seconds_________________________________________________________________________________
Problem 6 (6.13)
SOLUTIONS:
a) P(21.9 < X< 22.00) = P( 20.4 < Z< 0.4) = 0.3446b) P(21.9 < X< 22.01) = P( 20.4 < Z< 1.6) = 0.9452c) P(X> A) = 0.02 Z= 2.05 A = 22.0123
_________________________________________________________________________________
Section 6.3
Problem 7 (6.17)
SOLUTIONS:
a) Office I: X = 2.214 S= 1.718
Five-number summary 0.52 0.93 1.54 3.93 6.32Interquartile range = 3 1.33 S= 2.285Range = 5.8 6 S= 10.308
Between 1 XX S = 80%
Between 1.28 XX S = 90%
Between 2 XX S = 95%
The median is greater than the mean. The interquartile range isslightly larger than 1.33 times the standard deviation and therange is much smaller than 6 times the standard deviation. Thedistribution is right-skewed.
Note: The quartiles are obtained using Excel without anyinterpolation.
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Office I
Office II
Office II: X = 2.011 S= 1.892Five-number summary 0.08 0.60 1.505 3.75 7.55Interquartile range = 3.15 1.33 S= 2.516Range = 7.47 6 S= 11.350
Between 1 XX S = 75%
Between 1.28 XX S = 95%
Between 2 XX S = 95%
The median is smaller than the mean. The interquartile range islarger than 1.33 times the standard deviation and the range is muchsmaller than 6 times the standard deviation.The distribution is right-skewed.
Note: The quartiles are obtained using PHStat without any interpolation.
b)
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_________________________________________________________________________________
Problem 8 (6.19)Credit scores are three-digit numbers bylenders when evaluating your creditworthiness. The scores for residents of twentymetropolitan areas are as follows:
Decide weather or not the data appear to beapproximately normally distributed by:a. evaluating the actual versus the theoreticalproperties
b. constructing a normal probability plotc. constructing histogram
City Credit Score
Atlanta 670
Boston 705
Chicago 680
Cleveland 690
Dallas 653
Denver 675
Detroit 675
Houston 655Los Angeles 667
Miami 672
Minneapolis 707
New York 688
Orlando 671
Philadelphia 688
Phoenix 660
Sacramento 676
San Francisco 686
Seattle 691
Tampa 675
Washington 693
SOLUTIONS:
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Section 6.4
Problem 9 (6.23)
SOLUTIONS:a) P(5 < X< 7) = (7 5)/10 = 0.2
b) P(2 < X< 3) = (3 2)/10 = 0.1
c)0 10
52
+
= = d) ( )2
10 02.8868
12
= =
Problem 10 (6.27)
SOLUTIONS:a) P(X< 70) = (70 64)/(74 64) = 0.6b) P(65 < X< 70) = (70 65)/(74 64) = 0.5
c) P(X> 65) = (74 65)/(74 64) = 0.9
d)( )64+74
= 692
=( )
274 64
2.886812
= =
_________________________________________________________________________________
Section 6.5
Problem 11 (6.31)
SOLUTIONS:
a) P(arrival time 0.05) = 1 e ( 50)( 0.05)
= 0.9179
b) P(arrival time 0.0167) = 1 0.4339 = 0.5661c) If = 60,P(arrival time 0.05) = 0.9502,
P(arrival time 0.0167) = 0.6329d) If = 30,P(arrival time 0.05) = 0.7769
P(arrival time 0.0167) = 0.3941_________________________________________________________________________________
Problem 12 (6.33)
SOLUTIONS:
a) P(arrival time 0.05) = 1 e(15)(0.05)
= 0.5276
b) P(arrival time 0.25) = 0.9765_________________________________________________________________________________
Problem 13 (6.35)
SOLUTIONS:
a) P(X < 14) = (1/ 20)(14)1 0.5034e= =
b) P(X> 21) = ( )(1/ 20)(21)1 1 0.3499e= =c) P(X < 7) = (1/20)(7)1 0.2953e= =
Section 6.6
Problem 15 (6.41)
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SOLUTIONS:n= 90, p= 0.3333. np= 30 5 and n(1 p) = 60 5
np = = 30 ( )1np p = = 4.4721
a) P(X 20) P(X 19.5) = P(Z 2.3479) = 0.9906
b) P(X= 20)
P(19.5X
20.5) = P( 2.3479Z
2.1243) = 0.0074c) P(X