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Solved Consider that X is a base salary that a brand manager in Houston has Consider that Y a base salary that a brand manager in Los Angeles has In summary of all the details we can cle

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Hà Nội, tháng 12, năm 2022

29 An Internal Revenue Oversight Board survey found that 82% of taxpayers said

that it was very important for the Internal Revenue Service (IRS) to ensure that high-income tax payers do not cheat on their tax returns (The Wall Street Journal, February 11, 2009)

a, For a sample of eight taxpayers, what is the probability that at least six taxpayers say that it is very important to ensure that high-income tax payers do not cheat on their tax returns? Use the binomial distribution probability function shown in Section 5.4 to answer this question

For a binomial random variable X with parameters and the probability mass, function is , for

Let X = the number of taxpayers who say that it is very important to ensure that hight-income taxpayers do not cheat on their tax return

Then X is a binomial random variable with p=0.82 and n=8 The probability that

at least six taxpayers say that it is very important to ensure that hight-income taxpayers

do not cheat on their tax return is computed as:

= 0.2758 + 0.3590 + 0.2044

= 0.8392

b, For a sample of 80 taxpayers, what is the probability that at least 60 taxpayers say that it is very important to ensure that high-income tax payers do not cheat on their tax returns? Use the normal approximation of the binomial distribution to answer this question

X is a binomial random variable with and When using the normal approximation to the binomial, we set and in the definition of the normal curve

So we have:

The probability that at least 60 taxpayers say that it is very important to ensure that hight-income taxpayers do not cheat on their tax return is computed as:

c, As the number of trails in a binomial distribution application becomes large, what

is the advantage of using the normal approximation of the binomial distribution to compute probabilities?

The larger the number of observations in the sample, the more tedious it is to compute the exact probabilities of success by use of Equation (section a) Fortunately, though, when-ever the sample size is large, the normal distribution can be used to approximate the exact probabil-ities of success that otherwise would have to be obtained through laborious computations

30 When you sign up for a credit card, do you read the contract carefully? In a FindLaw.com survey, individuals were asked, “How closely do you read a contract for a credit card?” (USA Today, October 16, 2003) The findings were that 44% read every word, 33% read enough to understand the contract, 11% just glance at it, and 4% don’t read it at all

a For a sample of 500 people, how many would you expect to say that they read every word of a credit card contract?

Propotion of individuals who read every word of a credit card contract = 0.44 Size of the sample

Let X = people who read every word

So the expected number of people who read every word:

b For a sample of 500 people, what is the probability that 200 or fewer will say they read

every word of a credit card contract?

Given

So we have: (section a)

The probability that 200 or fewer will say they read every word of a credit card contract is computed as:

3

c For a sample of 500 people, what is the probability that at least 15 say they don’t read

credit card contracts?

Let Y = people who don’t read credit card contract

Given

So we have:

31 The probability that 15 or fewer will say they read every word of a credit card contract is computed as:

A Myrtle Beach resort hotel has 120 rooms In the spring months, hotel room occupancy is approximately 75%

Hotel rooms occupancy is approximately 75%,

Size of the sample

Let X = the number of rooms are occupied on a given day

So we have:

The z-score is the value (using the continuity correction) decreased by the mean

np and divided by the standard deviation

a, The probability that at least half of the rooms are occupied on a given day

b, The probability that 100 or more rooms are occupied on a given day

c, The probability that 80 or fewer rooms are occupied on a given day

46 Assume that the test scores from a college admissions tests are normally distributes, with a mean of 450 and a standard deviation of 100

a What percentage of the people taking the test score between 400 and 500?

b Suppose someone receives a score of 630 What percentage of the people taking the test score better? What percentage score worse?

c If a particular university will not admit anyone scoring below 480, what percentage

of the person taking the test would be acceptable to the university?

Sovled

Consider X as score in the test

which is based on the formula through which we have

where we have According to the details above we have:

a the percentage of people taking the test score between 400 and 500 is equal to the following probability according to normal distribution we have:

The value was obtained from the table of the Standard Normal Distribution We conclude that the percentage is approximately equal to 38%

b According to the details we have

The percentage of the people taking the test score worse than 630 is equal to the following probability according to normal distribution we have:

So the percentage is approximately equal to 96% We conclude that the percentage of the people taking the test score better is equal to

c We have that the percentage of the persons who would be acceptable is equal to the number of people with score better than 480 It is equal to:

5

So the percentage of the persons taking the test would be acceptable to the university

is approximately equal to 38%

47 According to Salary Wizard, the average base salary for a brand manager in Houston, Texas, is $88,592 and the average base salary for a brand manager in Los Angeles, California, is $97,417 (Salary Wizard website, February 27, 2008) Assume that salaries are normally dis-tributed, the standard deviation for brand managers in Houston is $19,900, and the standard deviation for brand managers in Los Angeles is $21,800

a What is the probability that a brand manager in Houston has a base salary in excess

of $100,000?

b What is the probability that a brand manager in Los Angeles has a base salary in ex-cess of $100,000?

c What is the probability that a brand manager in Los Angeles has a base salary of less than $75,000?

d How much would a brand manager in Los Angeles have to make in order to have a higher salary than 99% of the brand manger in Houston?

Solved

Consider that X is a base salary that a brand manager in Houston has

Consider that Y a base salary that a brand manager in Los Angeles has

In summary of all the details we can clearly see that:

The value set of salaries of people from Houston, Texas:

Set the value of the salary of managers from Los Angeles:

a, The probability that a brand manager in Houston has a base salary in excess of

$100,000 According to the formula, we have:

Overall, the probability that a brand manager in Houston has a base salary in excess

of $100,000 is 0,2832

b, The probability that a brand manager in Los Angeles has a base salary in excess of

$100,000, according to normal distribution:

We can conclude that the probability that a brand manager in Los Angeles has a base salary in excess of $100,000 is equal to 0,4528

c, The probability that a brand manager in Los Angeles has a base salary of less than

$75,000, according to normal distribution:

We can coclude that the probability that a brand manager in Los Angeles has a base salary of less than $75,000 is 0,1519

d, According the detail, we know that:

A brand manager in Los Angeles have to make in order to have a higher salary than 99% of the brand manger in Houston

So base on the normal distribution formula:

So we can conclude that the a brand manager in Los Angeles have to make $134959

in order to have a higher salary than 99% of the brand manger in Houston

48 A machine fills containers with a particular product A standard deviation of filling weights is known for past data to be 6 ounce If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must equal ? Assume the filling weights have a normal distribution

Sovled

Consider X is the mean filling weight for the machine

According to the details above we have:

We are given the following information in the question:

Based on the formular so

We are given that the weights is a bell shaped distribution that is a normal distribution:

7

Based on the information in the question, According to the normal distribution:

We can conclude that the mean of weight filling is 30.3

49 Consider a multiple-choice examination with 50 questions Each question has four possible answers Assume that a student who has done the homework and attended lectures hasa75% probability of answering any question correctly

a A student must answer 43 or more questions correctly to obtain a grade of A What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination? 36.2 %

b A student who answers 35 to 39 questions correctly will receive a grade of C What percentage of students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination? 48.077 %.since

c A student must answer 30 or more questions correctly to pass the examination What

percentage of the students who have done their homework and attended lectures will pass the examination? 99.3 % Since

d Assume that a student has not attended class and has not done the homework for the

course Furthermore, assume that the student will simply guess at the answer to each question What is the probability that this student will answer 30 or more questions correctly and pass the examination? reaches zero since

Solved

(*) In summary of all the details we can clearly see that :

Total questions that the examination have are

And students who have done the homework and attended lectures has a 75% probability of answering any question correctly 

So the mean questions that the students who have done the homework and attended lectures do correctly is ( questions)

The standard diviation is σ = = ~ 3.062

Consider X as the questions that students who have done the homework and attended lectures , do correctly Condition of X is ( X belongs to Natural numbers and

0 ≤X≤50)

These only apply to A ,B and C

a, To have the A score , a student must answer 43 or more questions correctly so to calculate probability of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination according to normal distribution we have

Percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination is 36.2 %

b, To have B score , A student who answers 35 to 39 questions correctly so to calculate probability of the students who have done their homework and attended lectures will obtain a grade of B on this multiple-choice examination according to normal distribution we have

Percentage of the students who have done their homework and attended lectures will obtain a grade of B on this multiple-choice examination is 48.077 %.since

c, To have C score a student must answer 30 or more questions correctly so to calculate probability of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination according to normal distribution we have

Percentage of the students who have done their homework and attended lectures will obtain a grade of C on this multiple-choice examination is 99.3 % Since

9

d, Consider x as total questions that the students who haven’t done the homework

nor attended lectures can do correctly (X belongs to Natural Numbers and 0 ≤X≤50 )

The mean questions that the students who haven’t done the homework nor attended lectures can do correctly is

Probability that student who haven’t done the homework nor attended lectures will answer 30 or more questions correctly and pass the examination is

So the probability that student who haven’t done the homework nor attended lectures will answer 30 or more questions correctly and pass the examination reaches zero since

50 A blackjack player at a Las Vegas casino learned that the house will provide a free room

if play is for four hours at an average bet of $50 The player’s strategy provides a probability of 49 of winning on any one hand, and the player knows that there are 60 hands per hour Suppose the player plays for four hours at a bet of $50 per hand

a What is the player’s expected payoff?

b What is the probability the player loses $1000 or more?

c What is the probability the player wins?

d Suppose the player starts with $1500 What is the probability of going broke?

Sovled

a, There are 60 hands per hour and the player plays for four hours then Total hands are :

The player’s strategy provides a probability of 49 of winning on any one hand so the probability of success is:

So propability of failure is

Also, following the above details we can see that since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand

So expected payoff on any one hand is

Since and expected payoff is -1$ per hand then

the expected payoff in four hours is:

(*) Note : The expected averange of total hands that player can win are :

Standard deviation of total hands are :

Consider X as total hands that player wins (X belongs to natural numbers and ) Normal distribution

b, According to the details we have above, the player may loose 1000 $ or more

means he has to loose at more than hands compared to those hands the player wins And

total hands are Which means He won less than 110 hands and lose more than 130 hands when he loose 1000 $ or more

The probability the player won less than 110 hands is

According to normal distribution : With and σ = 7.744

probability the player loses $1000 or more is 0.1632

c, If the player want to win , total hands he won must equal or more compared to hands he lose which means that player have to win 120 hands and more

The probability the player won more than 120 hands is

11