Probability &statistics

Question

News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we finished high school or college. And, in fact, only 20% of adults say they do not suffer from weight-loss woes. Suppose that the 20% figure is correct, and that a random sample of n = 120 adults is selected.

a. Does the distribution of p, the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation?

b. What is the probability that the sample proportion p exceeds .25?

c. What is the probability that p lies within the interval .25 to 30?

d. What might you conclude about p if the sample proportion exceeded .30?

Answer 1

a. Yes

np = 120(0.2) = 24

n(1-p) = 120(0.8) = 96

Both are greater than 10. The large counts condition tells us that p, the proportion of adults who do not suffer from excess weight, has an approximately normal distribution.

b. the mean and standard deviation of the sampling distribution are µ=0.2, σ=sqrt((0.2*0.8)/120)= 0.0365

P(p>0.25) = P(Z>((0.25 - 0.2)/0.0365) = P(Z>1.3699) = 0.0854

or

normalcdf(0.25, 1000, 0.02, 0.0365) = 0.0854

c. P(0.25 < p < 0.3) = P(p<0.30) - P(P<0.25)

P(p<0.3) = P(Z<((0.3 - 0.2)/.0365) = P(Z<2.7397) = 0.9969

P(0.25 < p < 0.3) = 0.9969 - (1-0.0854) = 0.0823

or

normalcdf(0.25, 0.3, 0.2, 0.0365) = 0.0823

d. P(p>0.30) = 1-0.9969 = 0.0031

If the sample proportion exceeds 0.30, I would conclude that the true proportion of adults who do not suffer from excess weight is not equal to 0.2. It is probably something larger.

Please let me know if you have any questions or if you would like a more detailed answer.