Share
Explore BrainMass

Application of Probability

1. Management at the local Randy Rooster Chicken Hut noted 50 percent of the customers order the crispy coating and the remainder order original coating.
a. What kind of probability distribution will you use to solve this problem?
b. What is the probability that none of the next four customers will request crispy?
c. What is the probability that at least two of the next four customers will request crispy?
d. What is the probability that no more than two will request crispy?
e. Construct the probability distribution function table and graph the distribution with a column chart. Comment on the shape of the graph.

2. Fifty faculty travel expense vouchers were submitted to the financial office during the summer. Of these 20 contained errors. If five vouchers are selected at random what is the probability that:
a. None contain errors.
b. One contains an error
c. Three or more contain errors
d. What kind of probability distribution did you use to solve this problem?
e. Make a graph of the PDF

3. Textbook authors and editors work very hard to minimize the number of errors in a text. However, they are not perfect and it is estimated that there are an average of 0.8 errors per chapter in a typical text book.
a. What kind of probability distribution will you use to solve this problem?
b. What is the probability that there are two or less errors in a particular textbook chapter?

4. Lunch customers arrive at Noodles and You at an average rate of 2.8 per minute. For each calculate the probability:
a. What kind of probability distribution will you use?
b. What is the probability of exactly five customers arriving in a minute?
c. What is the probability that no more than five customers will arrive in a given minute?
d. Create the graph of the PDF.

5. Bud and Lou's Construction build garages. They find is takes an average of 32 hours to build a garage with a standard deviation of two hours. Assuming that the time to construct follows a normal distribution:
a. What percent of the garages take from 32 to 34 hours to erect?
b. What percent of the garages take from 29 to 34 hours to erect?
c. What percent of the garages take 28.7 hours or less to erect?
d. Of all the garages, five (5) percent take how many hours or more to erect?

6. The annual commissions earned by the sales staff of SaleCo follows a normal distribution. The mean yearly commission is $40,000 with a standard deviation of $5000.
a. What percent of the sales reps earn more than $42,000 per year?
b. What percent of the sales reps earn between $32,000 and $42,000 per year?
c. What percent of the sales reps earn between $32,000 and $35,000 per year?
d. The sales manager wants to award the top 20 percent of the sales reps a bonus. What is the cutoff point for awarding those bonuses?

7. The fracture strength of a certain type of glass is normally distributed with a mean of 579 MPa and a standard deviation of 14 MPA:
a. What is the probability that a randomly chosen piece of glass will break at less than 579 BPa?
b. More than 590 BPa?
c. Less than 600 BPA?
d. Between 550 and 600 BPA?

8. In a quality test a sample of 100 nuts from a can of mixed nuts found that 19 were almonds.
a. Construct a 90 percent confidence interval for the true proportion of almonds.
b. May normality be assumed? Explain.
c. Using the proper formula for minimum sample size for a proportion, what sample size would be needed for a 90 percent confidence interval and an error of +/- .03?
d. Why does a quality control manager need to understand sampling?

9. You are asked to determine the typical number of sales of a certain product at the store you work for. A sample of 50 days over the last quarter shows an average of 55 units sold per day with a standard deviation of 10 units.
a. Construct a 95 percent confidence interval around the mean.
b. Construct a 99 percent confidence interval around the mean.
c. Show the two limits on a graph of some sort
d. Interpret your results.

10. Assume instead you selected a sample of 90 days sales. And that the mean and standard deviation were the same at 55 and 10 respectively.
a. Calculate the new 95 and 99 percent confidence intervals.
b. Show these on a graph of some sort.
c. Explain why the new intervals are narrower than the previous ones calculated.

11. A recent survey showed 4.6 percent of the sample of 250 had suffered some kind of identity theft in the past 12 months.
a. Construct a 90 percent confidence interval for the true proportion of those suffering identity theft.
b. Can we assume the distribution of the population is normal? Explain your answer.

12. A study showed the 14 of 180 publically traded business services companies failed a test for compliance with Sarbanes-Oxley requirements for financial records and fraud protection. Assuming these are a good random sample of all publically traded companies:
a. Construct a 90 percent confidence interval for the overall non-compliance proportion.
b. A 95 percent confidence interval.
c. A 99 percent confidence interval.
d. Explain why these differ.

13. The formula to calculate a minimum sample size for numerical data is as follows:

Where n is the sample size, z is the z value for the level of confidence chosen, s is the estimated standard deviation and E is the allowable error.
a. Using this formula calculate the minimum sample size for a study when the level of confidence is 95 percent, the standard deviation is $1000 and the allowable error is $100. What actual sample size might you suggest? Explain your answer.
b. How large of a sample size would be needed for a 99 percent level of confidence? What actual sample size might you suggest? Why is this sample size larger? Explain your answer.

Solution Preview

Please see the attachments.

The solution provides step by step method for the calculation so that you can understand the topic easily.

1. Management at the local Randy Rooster Chicken Hut noted 50 percent of the customers order the crispy coating and the remainder order original coating.

Answers

a. What kind of probability distribution will you use to solve this problem?

Binomial probability distribution

b. What is the probability that none of the next four customers will request crispy?

Let X be the number of customers order the crispy coating. Clearly X is binomial with n = 4 and p = 0.50. The probability mass function binomial variable is given by .The probability for different values of x are given below.

X P(X) P(<=X) P(<X) P(>X) P(>=X)
0 0.0625 0.0625 0 0.9375 1
1 0.25 0.3125 0.0625 0.6875 0.9375
2 0.375 0.6875 0.3125 0.3125 0.6875
3 0.25 0.9375 0.6875 0.0625 0.3125
4 0.0625 1 0.9375 0 0.0625

P (none of the next four customers will request crispy) = P (X = 0)

= 0.0625

c. What is the probability that at least two of the next four customers will request crispy?

P (at least two of the next four customers will request crispy) = P (X ≥ 2)

= 0.6875

d. What is the probability that no more than two will request crispy?

P (no more than two will request crispy) = P (X ≤ 2)

= 0.6875

e. Construct the probability distribution function table and graph the distribution with a column chart. Comment on the shape of the graph.

X P(X)
0 0.0625
1 0.25
2 0.375
3 0.25
4 0.0625

The distribution is symmetric.

2. Fifty faculty travel expense vouchers were submitted to the financial office during the summer. Of these 20 contained errors. If five vouchers are selected at random what is the probability that:

Answers

Let X be the number of faculty travel expense vouchers containing errors. Clearly X is binomial with n = 5 and p = 0.40. The probability mass function binomial variable is given by .The probability for different values of x are given below.

P(X) P(<=X) P(<X) P(>X) P(>=X)
0 0.07776 0.07776 0 0.92224 1
1 0.2592 0.33696 0.07776 0.66304 0.92224
2 0.3456 0.68256 0.33696 0.31744 0.66304
3 0.2304 0.91296 0.68256 0.08704 0.31744
4 0.0768 0.98976 0.91296 0.01024 0.08704
5 0.01024 1 0.98976 0 0.01024

a. None contain errors.

P (None contain errors) = P (X = 0) = 0.07776

b. One contains an error

P (One contains an error) = P (X = 1) = 0.2592

c. Three or more contain errors

P (Three or more contain errors) = P (X ≥ 3) = 0.31744

d. What kind of probability distribution did you use to solve this problem?

Binomial probability distribution

e. Make a graph of the PDF

3. Textbook authors and editors work very hard to minimize the number of errors in a text. However, they are not perfect and it is estimated that there are an average of 0.8 errors per chapter in a typical text book.

Answers

Let X be the number of errors in the text. Given that X is Poisson with mean  = 0.80. The probability density function of Poisson random variable is given by The probabilities for different values of X are given below.

X P(X) P(<=X) P(<X) P(>X) P(>=X)
0 0.449329 0.449329 0.000000 0.550671 1.000000
1 0.359463 0.808792 0.449329 0.191208 0.550671
2 0.143785 0.952577 0.808792 0.047423 0.191208
3 0.038343 0.990920 0.952577 0.009080 0.047423
4 0.007669 0.998589 0.990920 0.001411 0.009080
5 0.001227 0.999816 0.998589 0.000184 0.001411
6 0.000164 0.999979 0.999816 0.000021 0.000184
7 0.000019 0.999998 0.999979 0.000002 0.000021
8 0.000002 1.000000 0.999998 0.000000 0.000002
9 0.000000 1.000000 1.000000 0.000000 0.000000
10 0.000000 1.000000 1.000000 0.000000 0.000000
11 0.000000 1.000000 1.000000 0.000000 0.000000
12 0.000000 1.000000 1.000000 0.000000 0.000000
13 0.000000 1.000000 1.000000 0.000000 0.000000
14 0.000000 1.000000 1.000000 0.000000 0.000000
15 0.000000 1.000000 1.000000 0.000000 0.000000
16 0.000000 1.000000 1.000000 0.000000 0.000000
17 0.000000 1.000000 1.000000 0.000000 0.000000
18 0.000000 1.000000 1.000000 0.000000 0.000000
19 0.000000 1.000000 1.000000 0.000000 0.000000
20 0.000000 1.000000 1.000000 0.000000 0.000000

a. What kind of probability distribution will you use to solve this problem?

Poisson probability distribution

b. What is the probability that there are two or less errors in a particular textbook chapter?

P (there are two or less errors in a particular textbook chapter) = P (X ≤ 2)

= 0.952577

4. Lunch customers arrive at Noodles and You at an average rate of 2.8 per minute. For each calculate the probability:

Answers

a. What kind of probability distribution will you use?

Poisson probability distribution

b. What is the probability ...

Solution Summary

The applications of probability are determined. The kind of probability distributions to be used is given.

$2.19