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Understanding Continuous Probability Distributions & Computing Probabilities, Study notes of Introduction to Business Management

Solutions to exercises related to continuous probability distributions, including understanding the difference between discrete and continuous random variables, computing probability values for uniform, normal, and exponential distributions, and finding expected values and variances. Students studying statistics or probability theory will find this document useful for understanding these concepts and preparing for exams.

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Chapter 6
Continuous Probability Distributions
Learning Objectives
1. Understand the difference between how probabilities are computed for discrete and continuous
random variables.
2. 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.
3. Be able to compute probabilities using a normal probability distribution. Understand the role of the
standard normal distribution in this process.
4. Be able to compute probabilities using an exponential probability distribution.
5. Understand the relationship between the Poisson and exponential probability distributions.
6 - 1
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Chapter 6

Continuous Probability Distributions

Learning Objectives

  1. Understand the difference between how probabilities are computed for discrete and continuous

random variables.

  1. 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.

  1. Be able to compute probabilities using a normal probability distribution. Understand the role of the

standard normal distribution in this process.

  1. Be able to compute probabilities using an exponential probability distribution.
  2. Understand the relationship between the Poisson and exponential probability distributions.

Chapter 6

Solutions:

  1. 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) =.

  1. 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.

  1. 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) =.

  1. 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

  1. 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.

  1. 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) =.

  1. = .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 ( zz -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.

  1. 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.

  1. 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.

  1. 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.

  1. 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 =.

  1. 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 =.

  1. 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 =.

  1. = .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

/

  • e

/

= .6065 - .2231 = .38) = .10(6) =.

  1. 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

  1. 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.

  1. 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.

  1. 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.

  1. 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.

  1. 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) =.

  1. = .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.

  1. = .10(6) = .60.21% are acceptable.
  1. 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.