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Solutions

Solutions

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

24.

0.6

26.

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

28.

six

30.
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 5.43
33.

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

35.

0.5 to 9.5

37.

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

39.

μ = 5

41.
  1. Answers may vary.
  2. 3.5 7 3.5 7
43.
  1. Answers may vary.
  2. k = 7.25
  3. 7.25
45.

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.

47.

five

49.

f(x) = 0.2e-0.2x

51.

0.5350

53.

6.02

55.

f(x) = 0.75e-0.75x

57.
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 5.44
59.

0.4756

61.

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.

63.

continuous

65.

m = 0.000121

67.
  1. Answers may vary.
  2. P(x < 5,730) = 0.5001
69.
  1. Answers may vary.
  2. k = 2947.73
71.

Age is a measurement, regardless of the accuracy used.

73.
  1. Answers may vary.
  2. f(x)= 1 8 f(x)= 1 8 where 1≤x≤9 1≤x≤9
  3. five
  4. 2.3
  5. 15 32 15 32
  6. 333 800 333 800
  7. 2 3 2 3
75.
  1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
  2. Graph the probability distribution.
  3. f(x)= 1 8 f(x)= 1 8 where 0 ≤ x ≤ 8
  4. four
  5. 2.31
  6. 1 8 1 8
  7. 1 8 1 8
77.

d

78.

b

80.
  1. The probability density function of X is 1 25−16 = 1 9 1 25−16 = 1 9 .
    P(X > 19) = (25 – 19) ( 1 9 ) ( 1 9 ) = 6 9 6 9 = 2 3 2 3 .
    This shows the graph of the function f(x) = 1/9, the pdf for a uniform distribution. A horizontal line ranges from the point (16, 1/9) to the point (25, 1.9). Vertical lines extend from the x-axis to the graph at x = 16 and x = 25 creating a rectangle. A region is shaded inside the rectangle from x = 19 to x = 25. Text notes that the shaded area represents P(x > 19) = 2/3.
    Figure 5.45
  2. P(19 < X < 22) = (22 – 19) ( 1 9 ) ( 1 9 ) = 3 9 3 9 = 1 3 1 3 .
    This shows the graph of the function f(x) = 1/9, the pdf for a uniform distribution. A horizontal line ranges from the point (16, 1/9) to the point (25, 1.9). Vertical lines extend from the x-axis to the graph at x = 16 and x = 25 creating a rectangle. A region is shaded inside the rectangle from x = 19 to x = 22. Text notes that the shaded area represents P(19< x < 22) = 1/3.
    Figure 5.46
  3. 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 (25−18) 1 (25−18) = 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∩x>18) P(x>18) P(x>21∩x>18) P(x>18)
      = P(x>21) P(x>18) P(x>21) P(x>18) = (25−21) (25−18) (25−21) (25−18) = 4 7 4 7 .
82.
  1. P(X > 650) = 700−650 700−300 = 50 400 = 1 8 700−650 700−300 = 50 400 = 1 8 = 0.125
  2. P(400 < X < 650) = 650−400 700−300 = 250 400 650−400 700−300 = 250 400 = 0.625
84.
  1. X = the useful life of a particular car battery, measured in months.
  2. X is continuous.
  3. 40 months
  4. 360 months
  5. 0.4066
  6. 14.27
86.
  1. X = the time (in years) after reaching age 60 that it takes an individual to retire
  2. X is continuous.
  3. five
  4. five
  5. Answers may vary.
  6. 0.1353
  7. before
  8. 18.3
88.

a

90.

c

92.

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

NOTE

You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 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.
94.
  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 positive 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, 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 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive 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 positive people is discrete instead of continuous.)

96.

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) = 1−P( T<15 )=1−( 1− e − 15 7 )≈ e − 15 7 ≈0.1173 1−P( 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.