Question

According to Reader's Digest, 42 percent of primary care doctors think their patients receive unnecessary medical care.

What is the probability that the sample proportion will be within ±0.03 of the population proportion? (Round your answer to four decimal places.)

What is the probability that the sample proportion will be within ±0.05 of the population proportion? (Round your answer to four decimal places.)

Suppose data made available through a health system tracker showed health expenditures were $10,348 per person in the United States. Use $10,348 as the population mean and suppose a survey research firm will take a sample of 100 people to investigate the nature of their health expenditures. Assume the population standard deviation is $2,500.

What is the probability the sample mean will be within ±$150 of the population mean? (Round your answer to four decimal places.)

What is the probability the sample mean will be greater than $12,200? (Round your answer to four decimal places.)

Answer 1

for  Reader's Digest question sample size is not given

2)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	10348
std deviation   =σ=	2500.000
sample size       =n=	100
std error=σx̅=σ/√n=	250.00

a)

robability the sample mean will be within ±$150 of the population mean:

probability =P(10198<X<10498)=P((10198-10348)/250)<Z<(10498-10348)/250)=P(-0.6<Z<0.6)=0.7257-0.2743=0.4514

b)

probability the sample mean will be greater than $12,200:

probability =P(X>12200)=P(Z>(12200-10348)/250)=P(Z>7.41)=1-P(Z<7.41)=1-1=0.0000