The standard deviation of the lengths of hospital stay is 64

The standard deviation of the lengths of hospital stay is 6.4 days. Use this information to answer questions (a) and (b). a. For the variable ?length of hospital stay,? determine the sampling distribution of the sample mean for samples of 80 patients. The sample standard deviation is sigmax- = .7155. (Round to four decimal places as needed.) b. Obtain the probability that the sampling error made in estimating the population mean length of stay by the mean length of stay of a sample of 80 patients will be at most 2 days. The probability is approximately [0]. (Round to four decimal places as needed.)

Solution

The standard deviation of the lengths of hospital stay is 6.4 days.
a)   the sample mean for sample of 80 patients.
the sample standard deviation is ?X=

sd/sqrt(N)

= 6.4/sqrt(80)

= 0.7155 (to four decimals)

b) Obtain the probability that the sampling error made in estimating the population mean lenght of stay by the mean length of stay of a sample of 80 patients will be at most 2 dys.
What is the probability is approximately

E = z*sd/sqrt(N)

2 = z*6.4/sqrt(80)

z = 2.7952

prob (-2.7952 < z < 2.7952).