Question

1. . Suppose that weight loss during a diet for a first year medical school student is approximately normally distributed with a mean of 12.7 pounds and a standard deviation of 4.2 pounds.

(a) What is the probability that a randomly selected medical school student will lose 15 pounds or more during their first year? Round your final answer to four decimal places and label your answer with appropriate probability notation.

(b) What is the probability that a random sample of 20 medical students lose an average of 15 pounds or more during their first year of medical year? Round your final answer to four decimal places and label your answer with appropriate probability notation.

2.The owner of a popular nail salon is trying to attract more customers. The owner randomly selects people based on their purchase for a chance to receive a discount. When a customer's bill is printed, a program in the cash register randomly determines whether the customer will receive a discount on the bill. The program was written to generate a discount with a probability of 0.2, that is, giving a discount to 20 percent of the bills in the long run. However, the owner is concerned that the program has a mistake that results in the program not generating the intended long-run proportion of 0.2.

The owner selected a random sample of 100 receipts and found that only 16 percent of them received discounts. The conditions for inference are met. Using the sample data collected by the owner, calculate a 95% confidence interval for the true proportion of bills that will receive a discount eventually. Round your final answer to two decimal places.

In your opinion does the confidence interval you found in part

(a) provide convincing statistical evidence that the program is not working as intended? Justify your answer.

Answer 1

1)  mean 12.7 pounds and a standard deviation 4.2 pounds.

a) P(X>15)

P ( X>15 )=P ( X−μ>15−12.7 )=P((X−μ)/σ>(15−12.7)/4.2)
Since Z=(x−μ)/σ and (15−12.7)/4.2=0.55 we have:
P ( X>15 )=P ( Z>0.55 )
Use the standard normal table to conclude that:
P (Z>0.55)=0.2912

b) For samples n=20

mean 12.7 pounds and a standard deviation 4.2/sqrt(20)= 4.2/4.47= 0.94 pounds.

P ( X>15 )=P ( X−μ>15−12.7 )=P ( X−μσ>15−12.70.94)
Since Z=(x−μ)/σ and (15−12.7)/0.94=2.45 we have:
P ( X>15 )=P ( Z>2.45 )
Use the standard normal table to conclude that:
P (Z>2.45)=0.0071

NOTE: I HAVE DONE THE FIRST QUESTION PLEASE RE POST THE SECOND . THANK YOU.