Chapter 6- Do the problems

What if you need to find the probability of the x > 50?

1 - P(x < 50)

If you are given a sample size, what do you do?

Divide the standard deviation by the square root of the sample size.

What happens if you see finding the probability of the percentage?

You have the mean = p, then SD = square root of [((p(1-p))/n].

When you see "use the normal approximation to the binomial" what do you do?

You use the probability given for p and the sample size, n, and calculate the mean and SD. Proceed as normal.

Assuming that the heights of college women are normally distributed with mean 65 inches and standard deviation 2.5 inches. a. What percentage of women are taller than 65 inches? b. What percentage of women are shorter than 65 inches? What percentage of women are between 62.5 inches and 67.5 inches? What percentage of women are between 60 inches and 70 inches?

a. 50% b. 50% c. 68% d. 95%

Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12-hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean of 85 and estimated standard deviation of 25 (based on information from Diagnostic Tests with Nursing Applications, edited by S. Loeb, Springhouse). A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed. a. What is the probability that, on a single test, x < 40? b. Suppose a doctor uses the sample mean for two tests taken about a week apart. What can we say about the probability distribution of x-bar? What is the probability that x-bar < 40? (x-bar is the x with the line on top which means the mean of the sample.)

a. mean = 85, SD = 25 . P(x < 40) normcdf(neg inf, 40, 85, 25) b. mean = 85, SD = 17.68, P(x-bar < 40) normcdf(neg inf, 40, 85, 17.68)

Find the corresponding x-value for a probability to the left of .0195 with a mean of 20 and a standard deviation of 1.8.

invnorm(.0195, 20, 1.8) = 16.2845

Find the corresponding x-value for a probability to the right of .45 with a mean of 20 and a standard deviation of 1.8.

invnorm(.55, 20, 1.8) = 20.23

What would you use for each of the variables and probability? Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a Denver Post article, the mean of the x distribution is about $20 and the estimated standard deviation is about $7. Consider a random sample of 100 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. What is the probability that the mean of the sample is between $18 and $22?

mean = 20, SD = 0.7, P(18 < x < 22) normcdf(18, 22,20, 0.7)

One environmental group did a study of recycling habits in a California community. It found that 70% of the aluminum cans sold in the area were recycled. Using the normal approximation to the binomial, if 400 cans are sold today, what is the probability that 300 or more will be recycled?

mean = 280, SD = 2.898, P(x >= 300) = 1- P(x <= 300) 1 - normcdf(neg inf, 300, 280, 2.898)

The weights of Granny Smith apples from a large orchard are Normally distributed with a mean of 380 gm and a standard deviation of 28 gm. Three apples are selected at random from this orchard. What is the probability that their mean weight is greater than 400 gm.?

mean = 380, SD = 16.17, P(x-bar > 400) = 1 - P(x-bar < 400) 1 - normcdf(neg inf, 400, 380, 16.17)

The incomes in a certain large population of college teachers have a normal distribution with mean $60,000 and standard deviation $5000. Four teachers are selected at random from this population to serve on a salary review committee. What is the probability that their average salary exceeds $65,000?

mean = 60000, SD = 2500, P( x-bar > 65000) = 1 - P(x-bar < 65000) 1 - normcdf(neg inf, 65000, 60000, 2500)

What would you use for mean, standard deviation and the probability? The heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches (based on information from Statistical Abstract of the United States.) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall?

mean = 68, SD = 3, P(67 < x < 69) - There should be lines under the inequalities.) normcdf(67, 69, 68, 3)

What would you use for mean, standard deviation and the probability? Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of the coal loaded into each car are normally distributed, with a mean of 75 tons and standard deviation of 0.8 tons. What is the probability that one car chosen at random will have less than 74.5 tons of coal?

mean = 75, SD = 0.8, P(x < 74.5) You use these for the formula. normcdf(neg inf, 74.5, 5, 0.8)

A USA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 67% said that they drink it. Suppose that 70% of U.S. adults actually drink the cereal milk. Let p-hat be the proportion of people in the sample who drink the cereal milk. Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say they drink the cereal milk. Do you have any doubts about the result of this poll?

mean = p = .7, SD = . .010144, P(p-hat <.67) normcdf( neg inf, .67, .7, .010144)

Convert the x-value 15 to a z-value with mean of 12 and standard deviation of 1.5.

z = 2