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.
Figure 5.38
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. Check student’s solution.
2. $3.5 7 3.5 7$
43.
1. Check student's solution.
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.
Figure 5.39
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. Check student's solution
2. P(x < 5,730) = 0.5001
69.
1. Check student's solution.
2. k = 2947.73
71.

Age is a measurement, regardless of the accuracy used.

73.
1. Check student’s solution.
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) $( 1 9 ) ( 1 9 )$ = $6 9 6 9$ = $2 3 2 3$.
Figure 5.40
2. P(19 < X < 22) = (22 – 19) $( 1 9 ) ( 1 9 )$ = $3 9 3 9$ = $1 3 1 3$.
Figure 5.41
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. Check student’s solution.
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 TExp(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 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. 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 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.)

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.
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