# Uniform distribution, Normal Distribution

5. The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards (Golfweek,March 29,2003). Assume that the driving distance for these golfers is uniformly distributed over this interval.

a. Give a mathematical expression for the probability density function of driving distance.

b. What is the probability the driving distance for one of these golfers is less than 290 yards?

c. What is the probability the driving distance for one of these golfers is at least 300 yards?

d. What is the probability the driving distance for one of these golfers is between 290 and 305 yards?

e. How many of these golfers drive the ball at least 290 yards?

7. Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor's bid x is random variable that is uniformly distributed between $10,000 and $15,000.

a. Suppose you bid $12,000. What is the probability that you bid will accepted?

b. Suppose you bid $14,000. What is the probability that you bid will accepted?

c.What amount should you bid to maximize the probability that you get the property?

d. Suppose you know someone who is willing to pay you $16,000 for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

9. A random variable is normally distributed with a mean of µ=50 and a standard deviation of

a) a. Sketch a normal curve for the probability density function. Label the horizontal axis with values of 35,40,45,50,55,60 and 65. The figure below shows that the normal curve almost touches the horizontal axis at three standard deviations below and at three standard deviations above the mean (in this case at 35 and 65).

b) What is the probability the random variable will assume a value between 45 and 55?

c) What is the probability the random variable will assume a value between 40 and 60?

11. Given that z is a standard normal random variable, compute the following probabilities.

a) P(-1 <= z <= 0)

b) P( -1.5 <= z <= 0)

c) P( -2 <= z <= 0)

d) P(-2.5 <= z <= 0)

e) P( -3 <= z <= 0)

23. The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions.

a) What is the probability of completing the exam in one hour or less?

b) What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?

c) Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time?

https://brainmass.com/statistics/cumulative-distribution-function/uniform-distribution-normal-distribution-177399

#### Solution Preview

Please see attached file.

5. The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards (Golfweek,March 29,2003). Assume that the driving distance for these golfers is uniformly distributed over this interval.

a. Give a mathematical expression for the probability density function of driving distance.

f(x) = 1/(b-a) if a<=x<=b

0 otherwise

where

b= 310.6

a= 284.7

Thus

f(x) = 0.03861 if 284.7<=x<=310.6 =1/(310.6-284.7)

0 otherwise

b. What is the probability the driving distance for one of these golfers is less than 290 yards?

x< 290

Probability= 0.2046 =0.03861 x (290-284.7)

Answer: 0.2046

c. What is the probability the driving distance for one of these golfers is at least 300 yards?

x> 300

Probability= 0.4093 =0.03861 x (310.6-300)

Answer: 0.4093

d. What is the probability the driving distance for one of these golfers is between 290 and 305 yards?

290 <x< 300

Probability= 0.3861 =0.03861 x (300-290)

Answer: 0.3861

This can also be calculated using the values obtained in parts b and c

Probability (between 290 and 300 )= 1- Probability (less than 290)- Probability (more than 300)

= 0.3861 = 1-0.2046-0.4093

e. How many of these golfers drive the ball at least 290 yards?

Probability (greater than 290) = 1- Probability (less than 290)

Probability (less than 290)= 0.2046 (calculated in part b above)

Therefore , Probability (greater than 290) = 0.7954

n= 100 golfers

Therefore, number of golfers that drive the ball at least 290 yards= 80 =100x0.7954

(after rounding off)

Answer: 80

7. Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor's bid x is random variable that is uniformly distributed between $10,000 and $15,000.

f(x) = 1/(b-a) if a<=x<=b

0 otherwise

where

b= $15,000

a= $10,000

Thus

f(x) = 0.0002 if 10000<=x<=15000 =1/(15000-10000)

0 otherwise ...

#### Solution Summary

Questions on Uniform distribution, normal distribution have been answered.

Excel: binomial distribution, normal distribution, uniform

I need assistance solving these problem sets in excel format.

A large company must hire a new president. The Board of Directors prepares a list of five candidates, all of whom are equally qualified. Two of these candidates are members of a minority group. To avoid bias in the selection of the candidate, the company decides to select the president by lottery.

a. What is the probability one of the minority candidates is hired? (Round your answer to 1 decimal place.)

Probability

b. Which concept of probability did you use to make this estimate?

QUESTION 2

The chair of the board of directors says, "There is a 50% chance this company will earn a profit, a 30% chance it will break even, and a 20% chance it will lose money next quarter."

a. Use an addition rule to find the probability the company will not lose money next quarter. (Round your answer to 2 decimal places.)

=

b. Use the complement rule to find the probability it will not lose money next quarter. (Round your answer to 2 decimal places.)

=

QUESTION 3

A National Park Service survey of visitors to the Rocky Mountain region revealed that 50% visit Yellowstone Park, 40% visit the Tetons, and 35% visit both.

a. What is the probability a vacationer will visit at least one of these attractions? (Round your answer to 2 decimal places.)

Probability

b. What is the probability .35 called?

c. Are the events mutually exclusive?

QUESTION 4

P(A1) = .20, P(A2) = .40, and P(A3) = .40. P(B1|A1) = .25. P(B1|A2) = .05, and P(B1|A3) = .10.

Use Bayes' theorem to determine P(A3|B1). (Round your answer to 4 decimal places.)

P(A3|B1)

QUESTION 5

Solve the following:

a.

b.

9P 3

c.

7C 2

QUESTION 6

The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations.

a. What is the probability that all six arrive within 2 days? (Round your answer to 4 decimal places.)

Probability

b. What is the probability that exactly five arrive within 2 days? (Round your answer to 4 decimal places.)

Probability

c. Find the mean number of letters that will arrive within 2 days. (Round your answer to 1 decimal place.)

Number of letters

d-1. Compute the variance of the number that will arrive within 2 days. (Round your answer to 3 decimal places.)

Variance

d-2. Compute the standard deviation of the number that will arrive within 2 days. (Round your answer to 4 decimal places.)

Standard Deviation

QUESTION 7

The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations.

a. What is the probability that all six arrive within 2 days? (Round your answer to 4 decimal places.)

Probability

b. What is the probability that exactly five arrive within 2 days? (Round your answer to 4 decimal places.)

Probability

c. Find the mean number of letters that will arrive within 2 days. (Round your answer to 1 decimal place.)

Number of letters

d-1. Compute the variance of the number that will arrive within 2 days. (Round your answer to 3 decimal places.)

Variance

d-2. Compute the standard deviation of the number that will arrive within 2 days. (Round your answer to 4 decimal places.)

Standard Deviation

QUESTION 8

In a binomial distribution, n = 12 and π = .60.

a. Find the probability for x = 5? (Round your answer to 3 decimal places.)

Probability

b. Find the probability for x ≤ 5? (Round your answer to 3 decimal places.)

Probability

c. Find the probability for x ≥ 6? (Round your answer to 3 decimal places.)

Probability

QUESTION 9

A population consists of 15 items, 10 of which are acceptable.

In a sample of four items, what is the probability that exactly three are acceptable? Assume the samples are drawn without replacement. (Round your answer to 4 decimal places.)

Probability

QUESTION 10

According to the Insurance Institute of America, a family of four spends between $400 and $3,800 per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts.

a. What is the mean amount spent on insurance?

Mean $

b. What is the standard deviation of the amount spent? (Round your answer to 2 decimal places.)

Standard deviation $

c. If we select a family at random, what is the probability they spend less than $2,000 per year on insurance per year? (Round your answer to 4 decimal places.)

Probability

d. What is the probability a family spends more than $3,000 per year? (Round your answer to 4 decimal places.)

Probability

QUESTION 11

The mean of a normal probability distribution is 60; the standard deviation is 5. (Round your answers to 2 decimal places.)

a. About what percent of the observations lie between 55 and 65?

Percentage of observations %

b. About what percent of the observations lie between 50 and 70?

Percentage of observations %

c. About what percent of the observations lie between 45 and 75?

Percentage of observations %

QUESTION 12

A normal population has a mean of 12.2 and a standard deviation of 2.5.

a. Compute the z value associated with 14.3. (Round your answer to 2 decimal places.)

Z

b. What proportion of the population is between 12.2 and 14.3? (Round your answer to 4 decimal places.)

Proportion

c. What proportion of the population is less than 10.0? (Round your answer to 4 decimal places.)

Proportion

QUESTION 13

A normal population has a mean of 80.0 and a standard deviation of 14.0.

a. Compute the probability of a value between 75.0 and 90.0. (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.)

Probability

b. Compute the probability of a value of 75.0 or less. (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.)

Probability

c. Compute the probability of a value between 55.0 and 70.0. (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.)

Probability

QUESTION 14

For the most recent year available, the mean annual cost to attend a private university in the United States was $26,889. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,500.

Ninety-five percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number.)

Amount $

View Full Posting Details