Uniform Distribution

Normal Distribution

P(6 < x < 7)

one

zero

one

0.625

The probability is equal to the area from x = 32

$\frac{3}{2}$ to x = 4 above the x-axis and up to f(x) = 13

$\frac{1}{3}$.

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

1.5 ≤ x ≤ 4.5

0.3333

zero

0.6

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

six

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

0.5 to 9.5

f(x) = 19

$\frac{1}{9}$ where x is between 0.5 and 9.5, inclusive.

μ = 5

1. Check student’s solution.
2. 3.57
   $\frac{3.5}{7}$ 

1. Check student’s solution.
2. k = 7.25
3. 7.25

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.

five

f(x) = 0.2e^-0.2x

0.5350

6.02

f(x) = 0.75e^-0.75x

0.4756

The mean is larger. The mean is 1m=10.75≈1.33

$\frac{1}{�}=\frac{1}{0.75}\approx 1.33$, which is greater than 0.9242.

continuous

m = 0.000121

1. Check student’s solution
2. P(x < 5,730) = 0.5001

1. Check student’s solution.
2. k = 2947.73

Age is a measurement, regardless of the accuracy used.

1. Check student’s solution.
2. f(x)=18
   $�(�)=\frac{1}{8}$ 

   where 1≤x≤9 $1\le �\le 9$ 

3. five
4. 2.3
5. 1532
   $\frac{15}{32}$ 
6. 333800
   $\frac{333}{800}$ 
7. 23
   $\frac{2}{3}$ 

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)=18
   $�(�)=\frac{1}{8}$ 

   where 0 ≤ x ≤ 8

4. four
5. 2.31
6. 18
   $\frac{1}{8}$ 
7. 18
   $\frac{1}{8}$ 

d

b

1. The probability density function of X is 125−16=19
   $\frac{1}{25-16}=\frac{1}{9}$ 

   .
   P(X > 19) = (25 – 19) (19) $\left(\frac{1}{9}\right)$ 

   = 69 $\frac{6}{9}$ 

   = 23 $\frac{2}{3}$ 

   .

   

   Figure 5.40
2. P(19 < X < 22) = (22 – 19) (19)
   $\left(\frac{1}{9}\right)$ 

   = 39 $\frac{3}{9}$ 

   = 13 $\frac{1}{3}$ 

   .

   

   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)
     $\frac{1}{(25-18)}$ 

     = 17 $\frac{1}{7}$ 

     
     So, P(x > 21 | $|$ 

     x > 18) = (25 – 21)(17) $\left(\frac{1}{7}\right)$ 

     = 4/7.

   • Use the formula: P(x > 21 |
     $|$ 

     x > 18) = P(x>21∩x>18)P(x>18) $\frac{�(�>21\cap �>18)}{�(�>18)}$ 

     
     = P(x>21)P(x>18) $\frac{�(�>21)}{�(�>18)}$ 

     = (25−21)(25−18) $\frac{(25-21)}{(25-18)}$ 

     = 47 $\frac{4}{7}$ 

     .

1. P(X > 650) = 700−650700−300=50400=18
   $\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}$ 

   = 0.125

2. P(400 < X < 650) = 650−400700−300=250400
   $\frac{650-400}{700-300}=\frac{250}{400}$ 

   = 0.625

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

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

a

c

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) = 30e–30!

$\frac{3^0{�}^{–3}}{0\text{!}}$ = e^–3 ≈ 0.0498

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.

1. 1009
   $\frac{100}{9}$ 

   = 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 = 19
   $\frac{1}{9}$ 

   . The cumulative distribution function of X is P(X<x)=1−e−x9 $�\left(�<�\right)=1-{�}^{-\frac{�}{9}}$ 

   . Thus, P(X > 20) = 1 – P(X ≤ 20) = 1−(1−e−209)≈0.1084. $1-\left(1-{�}^{-\frac{20}{9}}\right)\approx \mathrm{0.1084.}$ 

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 = 17

$\frac{1}{7}$. The cdf is P(T < t) = 1−et7

$1-{�}^{\frac{�}{7}}$ 

1. P(T < 2) = 1 – 1−e−27
   $1-{�}^{-\frac{2}{7}}$ 

   ≈ 0.2485.

2. P(T > 15) = 1−P(T<15)=1−(1−e−157)≈e−157≈0.1173
   $1-�\left(�<15\right)=1-\left(1-{�}^{-\frac{15}{7}}\right)\approx {�}^{-\frac{15}{7}}\approx 0.1173$ 

   .

3. P(T > 15|T > 10) = P(T > 5) = 1−(1−e−57)=e−57≈0.4895
   $1-\left(1-{�}^{-\frac{5}{7}}\right)={�}^{-\frac{5}{7}}\approx 0.4895$ 

   .

4. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 307
   $\frac{30}{7}$ 

   , X ∼ Poisson(307) $\left(\frac{30}{7}\right)$ 

   . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311.