Introductory Statistics

# Solutions

Introductory 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.
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.
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 $1≤x≤9 1≤x≤9$
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. Graph the probability distribution.
4. $f ( x ) = 1 8 f ( x ) = 1 8$ where $0 ≤ x ≤ 8 0 ≤ x ≤ 8$
5. four
6. 2.31
7. $1 8 1 8$
8. $1 8 1 8$
9. 3.2
79.

d

81.

b

83.
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 5.54
2. P(19 < X < 22) = (22 – 19) $( 1 9 ) ( 1 9 )$ = $3 9 3 9$ = $1 3 1 3$.
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 $1 (25−18) 1 (25−18)$ = $1 7 1 7$
So, P(x > 21|x > 18) = (25 – 21)$( 1 7 ) ( 1 7 )$ = 4/7.
• Use the formula: P(x > 21|x > 18) =
= $P(x>21) P(x>18) P(x>21) P(x>18)$ = $(25−21) (25−18) (25−21) (25−18)$ = $4 7 4 7$.
85.
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
3. 0.10 = $width 700−300 width 700−300$, so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% 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

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
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 = .
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 $ln(area_to_the_right) (–m) = ln(1–0.2) – 1 8 ln(area_to_the_right) (–m) = ln(1–0.2) – 1 8$ = 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

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