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Chapter 6
Continuous Probability Distributions
Learning Objectives
- Understand the difference between how probabilities are computed for discrete and continuous
random variables.
- Know how to compute probability values for a continuous uniform probability distribution and be
able to compute the expected value and variance for such a distribution.
- Be able to compute probabilities using a normal probability distribution. Understand the role of the
standard normal distribution in this process.
- Be able to compute probabilities using an exponential probability distribution.
- Understand the relationship between the Poisson and exponential probability distributions.
Chapter 6
Solutions:
- a.
f ( x )
x
b. P ( x = 1.25) = 0. The probability of any single point is zero since the area under the curve above
any single point is zero.
c. P (1.0 x 1.25) = 2(.25) =.
d. P (1.20 < x < 1.5) = 2(.30) =.
- a.
f ( x )
x
b. P ( x < 15) = .10(5) =.
c. P (12 x 18) = .10(6) = .60) = .10(6) =.
d.
E x
e.
2
Var( ) 8) = .10(6) = .60.
x
Chapter 6
c. P ( x 255) = (0.045)(261.2 - 255) = 0.
d. P (245 x 260) = (0.045)(260 - 245) = 0.
e. P ( x 250) = 1 - P ( x < 250) = 1 - 0.4995 = 0.
The probability of anyone driving it 250 yards or more is 0.5005. With 60 players, the expected
number driving it 250 yards or more is (60)(0.5005) = 30.03. Rounding, I would expect 30 of
these women to drive the ball 250 yards or more.
- a. P (12 x 12.05) = .05(8) = .10(6) = .60) =.
b. P ( x 12.02) = .08) = .10(6) = .60(8) = .10(6) = .60) =.
c.
P x P x
Therefore, the probability is .04 + .64 = .68) = .10(6) =.
- a. P (10,000 x < 12,000) = 2000 (1 / 5000) =.
The probability your competitor will bid lower than you, and you get the bid, is .40.
b. P (10,000 x < 14,000) = 4000 (1 / 5000) = .8) = .10(6) =.
c. A bid of $15,000 gives a probability of 1 of getting the property.
d. Yes, the bid that maximizes expected profit is $13,000.
The probability of getting the property with a bid of $13,000 is
P (10,000 x < 13,000) = 3000 (1 / 5000) = .60.
The probability of not getting the property with a bid of $13,000 is .40.
The profit you will make if you get the property with a bid of $13,000 is $3000 = $16,000 -
13,000. So your expected profit with a bid of $13,000 is
EP ($13,000) = .6 ($3000) + .4 (0) = $18) = .10(6) = .6000.
If you bid $15,000 the probability of getting the bid is 1, but the profit if you do get the bid is only
$1000 = $16,000 - 15,000. So your expected profit with a bid of $15,000 is
EP ($15,000) = 1 ($1000) + 0 (0) = $1,000.
Continuous Probability Distributions
- a.
b. .68) = .10(6) = .6026 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50.
c. .9544 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50.
a..
b..
c..
d. .4938) = .10(6) =.
Continuous Probability Distributions
c. Look in the table for an area of .5000 - .0500 = .4500. Since .4500 is exactly halfway
between .4495 ( z = 1.64) and .4505 ( z = 1.65), we select z = 1.645. However, z = 1.64 or z =
1.65 are also acceptable answers.
d. Look in the table for an area of .5000 - .1000 = .4000. The area value in the table closest
to .4000 provides the value z = 1.28) = .10(6) = .60.
- Convert mean to inches: = 69
a. At x = 72
z =
P ( x 72) = 0.5000 + 0.3413 = 0.8) = .10(6) =.
P ( x > 72) = 1 - 0.8) = .10(6) = .60413 = 0.158) = .10(6) =.
b. At x = 60
z =
P ( x 60) = 0.5000 + 0.498) = .10(6) = .606 = 0.998) = .10(6) =.
P ( x < 60) = 1 - 0.998) = .10(6) = .606 = 0.
c. At x = 70
z =
P ( x 70) = 0.5000 + 0.1293 = 0.
At x = 66
z =
P ( x 66) = 0.5000 - 0.3413 = 0.158) = .10(6) =.
P (66 x 70) = P ( x 70) - P ( x 66) = 0.6293 - 0.158) = .10(6) = .607 = 0.
d. P ( x 72) = 1 - P ( x > 72) = 1 - 0.158) = .10(6) = .607 = 0.8) = .10(6) =.
- = .10(6) = .60. a. Find P ( x 60)
At x = 60
z =
P ( x < 60) = 0.5000 + 0.2549 = 0.
P ( x 60) = 1 - P ( x < 60) = 0.
b. Find P ( x 30)
At x = 30
z =
P ( x 30) = 0.5000 - 0.38) = .10(6) = .6030 = 0.
c. Find z -score so that P ( z z -score) = 0.
z -score = 1.28) = .10(6) = .60 cuts off 10% in upper tail
Now, solve for corresponding value of x.
Chapter 6
x
x = 49 + (16)(1.28) = .10(6) = .60) = 69.48) = .10(6) =.
So, 10% of subscribers spend 69.48) = .10(6) = .60 minutes or more reading The Wall Street Journal.
- We have = 3.5 and = .8) = .10(6) = .60.
a.
z
P ( x > 5.0) = P ( z > 1.8) = .10(6) = .608) = .10(6) = .60) = 1 - P ( z < 1.8) = .10(6) = .608) = .10(6) = .60) = 1 - .9699 =.
The rainfall exceeds 5 inches in 3.01% of the Aprils.
b.
z
P ( x < 3.0) = P ( z < -.63) = P ( z > .63) = 1 - P ( z < .63) = 1 - .7357 =.
The rainfall is less than 3 inches in 26.43% of the Aprils.
c. z = 1.28) = .10(6) = .60 cuts off approximately .10 in the upper tail of a normal distribution.
x = 3.5 + 1.28) = .10(6) = .60(.8) = .10(6) = .60) = 4.
If it rains 4.524 inches or more, April will be classified as extremely wet.
- We use = 27 and = 8) = .10(6) =.
a.
z
P ( x 11) = P ( z -2) = .5000 - .4772 = .0228) = .10(6) =.
The probability a randomly selected subscriber spends less than 11 hours on the computer is .025.
b.
z
P ( x > 40) = P ( z > 1.63) = 1 - P ( z 1.63) = 1 - .948) = .10(6) = .604 =.
5.16% of subscribers spend over 40 hours per week using the computer.
c. A z -value of .8) = .10(6) = .604 cuts off an area of .20 in the upper tail.
x = 27 + .8) = .10(6) = .604(8) = .10(6) = .60) = 33.
A subscriber who uses the computer 33.72 hours or more would be classified as a heavy user.
Chapter 6
Therefore 15.8) = .10(6) = .607% of students will not complete on time.
We would expect 9.522 students to be unable to complete the exam in time.
- a.
i
x
x
n
2
i
x x
s
n
We will use
x as an estimate of and s as an estimate of in parts (b) - (d) below.
b. Remember the data are in thousands of shares.
At 8) = .10(6) =.
z
P ( x 8) = .10(6) = .6000) = P ( z -.90) = 1 - P ( z .90) = 1 - .8) = .10(6) = .60159 = .18) = .10(6) =.
The probability trading volume will be less than 8) = .10(6) = .6000 million shares is .18) = .10(6) =.
c. At 1000
z
P ( x 1000) = P ( z .8) = .10(6) = .605) = 1 - P ( z .8) = .10(6) = .605) = 1 - .8) = .10(6) = .60023 =.
The probability trading volume will exceed 1 billion shares is.
d. A z -value of 1.645 cuts off an area of .05 in the upper tail
x = + z = 902.75 + 1.645(114.18) = .10(6) = .605) = 1,090.58) = .10(6) =.
They should issue a press release any time share volume exceeds 1,091 million.
- a. Find P ( x > 100)
At x = 100
z =
P ( x > 100) = P ( z .5) = 0.
b. Find P ( x 90)
At x = 90
z =
P ( x 90) = .5000 - .3413 = 0.158) = .10(6) =.
Continuous Probability Distributions
c. Find P (8) = .10(6) = .600 x 130)
At x = 130
z =
P ( x 130) = 0.8) = .10(6) =.
At x = 8) = .10(6) =.
z
Area to left is .0668) = .10(6) =.
P (8) = .10(6) = .600 x 130) = .8) = .10(6) = .60413 - .0668) = .10(6) = .60 =.
- a. P ( x 6) = 1 - e
-6/8) = .10(6) =.
b. P ( x 4) = 1 - e
-4/8) = .10(6) =.
c. P ( x 6) = 1 - P ( x 6) = 1 - .5276 =.
d. P (4 x 6) = P ( x 6) - P ( x 4) = .5276 - .3935 =.
- a.
0
/ 3
0
x
P x x e
b. P ( x 2) = 1 - e
-2/
c. P ( x 3) = 1 - P ( x 3) = 1 - (1 - e
3 / 3
) = e
d. P ( x 5) = 1 - e
-5/
e. P (2 x 5) = P ( x 5) - P ( x 2) = .8) = .10(6) = .60111 - .48) = .10(6) = .6066 =.
- = .10(6) = .60. a. P ( x 10) = 1 - e
-10/
b. P ( x 30) = 1 - P ( x 30) = 1 - (1 - e
-30/
) = e
-30/
c. P (10 x 30) = P ( x 30) - P ( x 10)
= (1 - e
/
) - (1 - e
/
= e
/
/
= .6065 - .2231 = .38) = .10(6) =.
- a.
Continuous Probability Distributions
then f(x) = 30 e
-30 x
b. A month is 1/12 of a year so,
30 /12 30 /
P x P x e e
The probability of no transaction during January is the same as the probability of no transaction
during any month: .08) = .10(6) =.
c. Since 1/2 month is 1/24 of a year, we compute,
30 / 24
P x e
- a. Let x = sales price ($1000s)
for 200 225
0 elsewhere
x
f x
b. P ( x 215) = (1 / 25) (225 - 215) = 0.
c. P ( x < 210) = (1 / 25)(210 - 200) = 0.
d. E ( x ) = (200 + 225)/2 = 212,
If she waits, her expected sale price will be $2,500 higher than if she sells it back to her company
now. However, there is a 0.40 probability that she will get less. It’s a close call. But, the
expected value approach to decision making would suggest she should wait.
- a. For a normal distribution, the mean and the median are equal.
b. Find the z -score that cuts off 10% in the lower tail.
z -score = -1.28) = .10(6) =.
Solving for x ,
x – 63,
x = 63,000 - 1.28) = .10(6) = .60 (15000)
The lower 10% of mortgage debt is $43,8) = .10(6) = .6000 or less.
c. Find P ( x > 8) = .10(6) = .600,000)
At x = 8) = .10(6) = .600,
Chapter 6
z =
P ( x > 8) = .10(6) = .600,000) = 1.0000 - .8) = .10(6) = .60708) = .10(6) = .60 = 0.
d. Find the z -score that cuts off 5% in the upper tail.
z -score = 1.645. Solve for x.
x – 63,
x = 63,000 + 1.645 (15,000)
The upper 5% of mortgage debt is in excess of $8) = .10(6) = .607,675.
- a. P ( defect ) = 1 - P (9.8) = .10(6) = .605 x 10.15)
= 1 - P (-1 z 1)
Expected number of defects = 1000(.3174) = 317.
b. P ( defect ) = 1 - P (9.8) = .10(6) = .605 x 10.15)
= 1 - P (-3 z 3)
Expected number of defects = 1000(.0028) = .10(6) = .60) = 2.8) = .10(6) =.
c. Reducing the process standard deviation causes a substantial reduction in the number of defects.
- a. At 11%, z = -1.
x –
Therefore , =
b.
z
Area to left is .5000 - .3255 =.
z
Area to left is.
Chapter 6
P ( x 8) = .10(6) = .600,000) = .5000 - .4406 = 0.
P (8) = .10(6) = .600,000 x 150,000) = 0.78) = .10(6) = .6023 - 0.0594 = 0.
b. Find P ( x < 50,000)
At x = 50,
z =
P ( x < 50,000) = .5000 - .4948) = .10(6) = .60 = 0.
c. Find the z -score cutting off 95% in the left tail.
z -score = 1.645. Solve for x.
x – 126,68) = .10(6) =.
x = 126,68) = .10(6) = .601 + 1.645 (30,000)
The probability is 0.95 that the number of lost jobs will not exceed 176,031.
- a. At 400,
z
Area to left is .308) = .10(6) =.
At 500,
z
Area to left is.
P (400 x 500) = .6915 - .308) = .10(6) = .605 = .38) = .10(6) =.
- = .10(6) = .60.3% will score between 400 and 500.
b. At 630,
z
96.41% do worse and 3.59% do better.
c. At 48) = .10(6) = .600,
z
Area to left is.
- = .10(6) = .60.21% are acceptable.
- a. At 75,
Continuous Probability Distributions
z
P ( x > 75,000) = P ( z > 1.14) = 1 - P ( z 1.14) = 1 - .8) = .10(6) = .60729 =.
The probability of a woman receiving a salary in excess of $75,000 is.
b. At 75,
z
P ( x > 75,000) = P ( z > 1.36) = 1 - P ( z 1.36) = 1 - .9131 = .08) = .10(6) =.
The probability of a man receiving a salary in excess of $75,000 is .08) = .10(6) =.
c. At x = 50,
z
P ( x < 50,000) = P ( z < -2.43) = 1 - P ( z < 2.43) = 1 - .9925 =.
The probability of a woman receiving a salary below $50,000 is very small:.
d. The answer to this is the male copywriter salary that cuts off an area of .01 in the upper tail of the
distribution for male copywriters.
Use z = 2.
x = 65,500 + 2.33(7,000) = 8) = .10(6) = .601,8) = .10(6) =.
A woman who makes $8) = .10(6) = .601,8) = .10(6) = .6010 or more will earn more than 99% of her male counterparts.
At 2%
z = -2.05 x = 18) = .10(6) =.
x
z
= 18) = .10(6) = .60 + 2.05 (.6) = 19.23 oz.
