# Statistics of ways Americans relieve stress

1) One of the ways most Americans relieve stress is to reward themselves with sweets. According to one study, 46% admit to overeating sweet foods when stressed. Suppose that the 46% figure is correct and that a random sample of n = 100 Americans is selected.

a) Does the distribution of p^, the sample proportion of Americans who relieve stress by overeating sweet foods, have an approximately normal distribution? If so, what are its mean and standard deviation?

b) What is the probability that the sample proportion p^, exceeds .5?

c) What is the probability that p^ lies within the interval .35 to .55?

d) What might you conclude if the sample proportion were as small as 30%?

2) The proportion of individuals with an Rh-positive blood type is 85%. You have a random sample of n = 500 individuals.

a) What are the mean and standard deviation of p^, the sample proportion with Rh-positive blood type?

b) Is the distribution of p^ approximately normal? Justify your answer.

c) What is the probability that the sample proportion p^ exceeds 82%?

d) What is the probability that the sample proportion lies between 83% and 88%?

e) 99% of the time, the sample proportion would lie between what two limits?

https://brainmass.com/statistics/probability/statistics-americans-relieve-stress-37735

#### Solution Preview

Statistics - Sample Problems 7A

1) One of the ways most Americans relieve stress is to reward themselves with sweets. According to one study, 46% admit to overeating sweet foods when stressed. Suppose that the 46% figure is correct and that a random sample of n = 100 Americans is selected.

a) Does the distribution of p^, the sample proportion of Americans who relieve stress by overeating sweet foods, have an approximately normal distribution? If so, what are its mean and standard deviation?

From the question, we know that the expected number of overeating sweet is 46% * 100 = 46 persons.

Then the mean of the distribution is p = 46/100 = 46%

The standard deviation is SE = SQRT(p*(1-p)/n) = SQRT(0.46*(1-0.46)/100)= 0.05

And the distribution of p is approximately normal due to the ...

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