ch6 continuous probability - suggested problems solutions

8/8/2019 Ch6 Continuous Probability - Suggested Problems Solutions

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STATISTICS FOR SOCIAL SCIENCESBUS 152

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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

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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

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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%


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 1.28 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

= =

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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

ch6 continuous probability - suggested problems solutions

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