Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Statistics

Solutions

StatisticsSolutions

1.

Uniform distribution

3.

Normal distribution

5.

P(6 < x < 7)

7.

one

9.

zero

11.

one

13.

0.625

15.

The probability is equal to the area from x = 3 2 3 2 to x = 4 above the x-axis and up to f(x) = 1 3 1 3 .

17.

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

19.

1.5 ≤ x ≤ 4.5

21.

0.3333

23.

zero

25.

0.6

27.

b is 12, and it represents the highest value of x.

29.

six

31.
This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.
Figure 5.52
33.

4.8

35.

X = The age (in years) of cars in the staff parking lot

37.

0.5 to 9.5

39.

f(x) = 1 9 1 9 where x is between 0.5 and 9.5, inclusive.

41.

μ = 5

43.
  1. Check student’s solution.
  2. 3.5 7 3.5 7
45.
  1. Check student's solution
  2. k = 7.25
  3. 7.25
47.

No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

49.

five

51.

f(x) = 0.2e–0.2x

53.

0.5350

55.

6.02

57.

f(x) = 0.75e–0.75x

59.
This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75.
Figure 5.53
61.

0.4756

63.

The mean is larger. The mean is 1 m = 1 0.75 1.33 1 m = 1 0.75 1.33 , which is greater than 0.9242.

65.

continuous

67.

m = 0.000121

69.
  1. Check student's solution
  2. P(x < 5,730) = 0.5001
71.
  1. Check student's solution
  2. k = 2947.73
73.

Age is a measurement, regardless of the accuracy used.

75.
  1. X ~ U(1, 9)
  2. Check student’s solution
  3. f(x)= 1 8 f(x)= 1 8 where 1x9 1x9
  4. five
  5. 2.3
  6. 15 32 15 32
  7. 333 800 333 800
  8. 2 3 2 3
  9. 8.2
77.
  1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
  2. X ~ U(0, 8)
  3. f(x)= 1 8 f(x)= 1 8 where ≤ x ≤ 8
  4. four
  5. 2.31
  6. 1 8 1 8
  7. 1 8 1 8
  8. 3.2
79.

d

81.

b

83.
  1. The probability density function of X is 1 2516 = 1 9 1 2516 = 1 9 .
    P(X > 19) = (25 – 19) ( 1 9 ) ( 1 9 ) = 6 9 6 9 = 2 3 2 3 .
    A graph is shown. The x($) axis shown is measured from 14 to 26 in increments of two. A box is drawn on the graph between the measurements of 16 and 25, and up to 1/9 on the y axis. The box is shaded blue from 19 to 25. Text above the box states “shaded area represents P(x>19)=2/3.
    Figure 5.54
  2. P(19 < X < 22) = (22 – 19) ( 1 9 ) ( 1 9 ) = 3 9 3 9 = 1 3 1 3 .
    A graph is shown. The x($) axis shown is measured from 14 to 26 in increments of two. A box is drawn on the graph between the measurements of 16 and 25 and up to 1/9 on the y axis. The box is shaded from 23 to 25. Text above the box states “Shaded area represents P(19<x<22)=1/3.
    Figure 5.55
  3. The area must be 0.25, and 0.25 = (width) ( 1 9 ) ( 1 9 ) , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
  4. This is a conditional probability question. P(x > 21| x > 18). You can do this two ways:
    • Draw the graph where a is now 18 and b is still 25. The height is 1 (2518) 1 (2518) = 1 7 1 7
      So, P(x > 21|x > 18) = (25 – 21) ( 1 7 ) ( 1 7 ) = 4/7.
    • Use the formula: P(x > 21|x > 18) = P(x>21 AND x>18) P(x>18) P(x>21 AND x>18) P(x>18)
      = P(x>21) P(x>18) P(x>21) P(x>18) = (2521) (2518) (2521) (2518) = 4 7 4 7 .
85.
  1. P(X > 650) = 700650 700300 = 50 400 = 1 8 700650 700300 = 50 400 = 1 8 = 0.125.
  2. P(400 < X < 650) = 650400 700300 = 250 400 650400 700300 = 250 400 = 0.625
  3. 0.10 = width 700300 width 700300 , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the farthest 10 percent of days.
87.
  1. X = the useful life of a particular car battery, measured in months.
  2. X is continuous.
  3. X ~ Exp(0.025)
  4. 40 months
  5. 360 months
  6. 0.4066
  7. 14.27
89.
  1. X = the time (in years) after reaching age 60 that it takes an individual to retire
  2. X is continuous.
  3. X ~ Exp ( 1 5 ) ( 1 5 )
  4. five
  5. five
  6. Check student’s solution.
  7. 0.1353
  8. before
  9. 18.3
91.

a

93.

c

95.

Let T = the life time of a light bulb.

The decay parameter is m = 1/8, and T ~ Exp(1/8). The cumulative distribution function is P(T<t)=1 e t 8 P(T<t)=1 e t 8

  1. Therefore, P(T < 1) = 1 – e 1 8 1 8 ≈ 0.1175.
  2. We want to find P(6 < t < 10).
    To do this, P(6 < t < 10) – P(t < 6)
    = =( 1 e 1 8 *10 )( 1 e 1 8 *6 ) =( 1 e 1 8 *10 )( 1 e 1 8 *6 ) ≈ 0.7135 – 0.5276 = 0.1859
    A line graph is shown. The t axis measures values from 0 to 60 in increments of 20. The y axis measures values from 0 to 0.12 in increments of 0.02. Line starts from t, y coordinates of 0, 0.12, and then takes a steep curve downwards before leveling out at the t,y coordinates of 60, 0. The segment of the graph from 6 to 10 on the t axis is shaded. Text above the graph reads “shaded area represents probability P(6<t<10)=0.1859”.
    Figure 5.56
  3. We want to find 0.70 =P(T>t)=1( 1 e t 8 )= e t 8 . =P(T>t)=1( 1 e t 8 )= e t 8 .
    Solving for t, e t 8 t 8 = 0.70, so t 8 t 8 = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years
    Or use t = ln(area_to_the_right) (m) = ln(0.70) 1 8 2.85 years ln(area_to_the_right) (m) = ln(0.70) 1 8 2.85 years .
    A line graph is shown. The t axis measures values from 0 to 40 in increments of 20. The y axis measures values from 0 to 0.12 in increments of 0.02. Line starts from t, y coordinates of 0, 0.12, and then takes a steep curve downwards before leveling out at the t,y coordinates of 40, 0. The segment of the graph from 2.85 to 40 on the t axis is shaded. Text above the graph reads “shaded area represents probability P(t>2.85)=0.70”.
    Figure 5.57
  4. We want to find 0.02 = P(T < t) = 1 – e t 8 t 8 .
    Solving for t, e t 8 t 8 = 0.98, so t 8 t 8 = ln(0.98), and t = –8ln(0.98) ≈ 0.1616 years, or roughly two months.
    The warranty should cover light bulbs that last less than 2 months.
    Or use ln(area_to_the_right) (m) = ln(10.2) 1 8 ln(area_to_the_right) (m) = ln(10.2) 1 8 = 0.1616.
  5. We must find P(T < 8|T > 7).
    Notice that by the rule of complement events, P(T < 8|T > 7) = 1 – P(T > 8|T > 7).
    By the memoryless property (P(X > r + t|X > r) = P(X > t)).
    So P(T > 8|T > 7) = P(T > 1) = 1( 1 e 1 8 )= e 1 8 0.8825 1( 1 e 1 8 )= e 1 8 0.8825
    Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175.
97.

Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3.
Therefore, (X = 0) = 3 0 e 3 0! 3 0 e 3 0! = e–3 ≈ 0.0498

You could let T = duration of time between no-hitters. Since the time is exponential and there are three no-hitters per season, then the time between no-hitters is 1 3 1 3 season. For the exponential, µ = 1 3 1 3 .
Therefore, m = 1 μ 1 μ = 3 and T ~ Exp(3).

  1. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e–3) = e–3 ≈ 0.0498.
  2. Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a.
  3. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528.
99.
  1. 100 9 100 9 = 11.11
  2. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.
  3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 1 9 . The cumulative distribution function of X is P( X<x )=1 e x 9 P( X<x )=1 e x 9 . Thus hus, P(X > 20) = 1 - P(X ≤ 20) = 1( 1 e 20 9 )0.1084. 1( 1 e 20 9 )0.1084.

Note

We could also deduce that each person arriving has a 8 9 8 9 chance of not having type B blood. So the probability that none of the first 20 people arrive have type B blood is ( 8 9 ) 20 0.0948 ( 8 9 ) 20 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between type B people is discrete instead of continuous.)

101.

Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 1 7 . The cdf is P(T < t) = 1 e t 7 1 e t 7

  1. P(T < 2) = 1 - 1 e 2 7 1 e 2 7 ≈ 0.2485.
  2. P(T > 15) = 1P( T<15 )=1( 1 e 15 7 ) e 15 7 0.1173 1P( T<15 )=1( 1 e 15 7 ) e 15 7 0.1173 .
  3. P(T > 15|T > 10) = P(T > 5) = 1( 1 e 5 7 )= e 5 7 0.4895 1( 1 e 5 7 )= e 5 7 0.4895 .
  4. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 30 7 , X ~ Poisson ( 30 7 ) ( 30 7 ) . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311.
Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-statistics . Changes were made to the original material, including updates to art, structure, and other content updates.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/statistics/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/statistics/pages/1-introduction
Citation information

© Jan 23, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.