Question

Students of a large university spend an average of $5 a day on lunch. The standard
deviation of the expenditure is $3. A simple random sample of 36 students is taken.
a. What are the expected value and standard deviation of the sampling distribution
of the sample mean?
b. What is the probability that the sample mean will be at least $4?
c. What is the probability that the sample mean will be at least $5.90?

Answer 1

Normal assumption, population mean=5, stdev=3, so var=3^2=9; s = sample of 36
a) expected mean=5, var=9/36, stdev=3/sqrt(36)=0.5, normal shape
b) prob(s<4) CDF @ (4-5)/0.5 (i.e. > -2*sigma) = 1-2.275% = 97.7%
c) prob(s>5.90) CDF @ (5.90-5)/0.5 (i.e.> +1.8*sigma) = 1-0.964 = 3.59%

Answer 2

Normal assumption, population mean=5, stdev=3, so var=3^2=9; s = sample of 36
a) expected mean=5, var=9/36, stdev=3/sqrt(36)=0.5, normal shape
b) prob(s<4) CDF @ (4-5)/0.5 (i.e. > -2*sigma) = 1-2.275% = 97.7%
c) prob(s>5.90) CDF @ (5.90-5)/0.5 (i.e.> +1.8*sigma) = 1-0.964 = 3.59%