# Minitab assignment 3 spring 2014

### Need your ASSIGNMENT done? Use our paper writing service to score better and meet your deadline.

Order a Similar Paper HERE Order a Different Paper HERE

You will be calculating the mean, standard deviation, and probabilities based on the binomial distribution and calculating probabilities based on the normal distribution. Use MINITAB to calculate these probabilities and to find the means and standard deviations. You will also generate a sampling distribution.

Make sure you do the following:

1. Put your name on your paper.

2. Under your name, put STAT 100 with your section number.

3. Put MINITAB Assignment 3 centered on top of Page 1.

4. Number and letter the answers accordingly.

5. Use complete and coherent sentences to answer the questions.

6. Copy and paste of all generated graphics into your paper.

7. Save file with your name and MINITAB Assignment 3.

8. Upload file on Canvas.

1. According to the Information Please Almanac, 80% of adult smokers started smoking before turning 18 years old. Suppose 10 smokers 18 years old or older are randomly selected and the number of smokers who started smoking before 18 is recorded. For parts a-c, copy and paste the MINITAB commands you used and the results found in the Session window. Use the cumulative probabilities found in part c to answer parts d-g. Round answers to 4 decimal places.

a. What is the mean of the number of smokers who started smoking before 18 years of age for a sample of 10 smokers?

b. What is the standard deviation of the number of smokers who started smoking before 18 years of age for a sample of 10 smokers?

c. What is the cumulative distribution function (CDF) of this binomial?

d. Find the probability that exactly 8 of the 10 smokers started smoking before 18 years of age.

e. Find the probability that at least 8 of the 10 smokers started smoking before 18 years of age.

f. Find the probability that fewer than 8 of the 10 smokers started smoking before 18 years of age.

g. Find the probability that between 7 and 9 (inclusive) of the 10 smokers started smoking before 18 years of age. (In other words, find the probability that 7, 8, or 9 of the 10 smokers started smoking before 18 years of age.)

2. General Electric manufactures a decorative Crystal Clear 60-watt light bulb that it advertises will last 1500 hours. Suppose the lifetimes of the light bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. Using MINITAB, find the following probabilities; round answers to 4 decimal places. For each part, copy and paste the commands you used and the results found in the Session window along with the answer.

a. What proportion of the light bulbs will last less than the advertised time (1500 hours)?

b. What proportion of the light bulbs will last more than 1650 hours?

c. What proportion of the light bulbs will last between 1625 and 1725 hours?

d. If a sample of 10 light bulbs is randomly selected, what is the probability the mean number of hours that the sample of 10 light bulbs will last is more than 1650 hours?

3. According to the AAA (American Automobile Association) in April 20, 2005, a family of two adults and two children on vacation in the United States will pay an average of $247.02 per day for food and lodging with a standard deviation of $60.41 per day.

For each part:

If the probability can be determined, copy and paste the commands you used and the results found in the Session window along with the answer.

If the probability cannot be determined, then answer: “Probability cannot be determined with information given.”

a. Suppose a random sample of 10 families of two adults and two children is selected and monitored while on vacation in the United States. What is the probability that the average daily expenses for the sample are over $250 per day?

b. Suppose a random sample of 50 families of two adults and two children is selected and monitored while on vacation in the United States. What is the probability that the average daily expenses for the sample are over $260 per day?