Find the interval within which 95 percent of the sample mean

Find the interval within which 95 percent of the sample means would be expected to fall, assuming that each sample is from a normal population. (a) p = 183. sigma = 20, n = 50. (Round your answers to 2 decimal places.) The 95% range is from to (b) p = 860, sigma = 12, n = 6. (Round your answers to 2 decimal places.) The 95% range is from to (c) p = 77, sigma= 4, n = 27. (Round your answers to 3 decimal places.) The 95% range is from to

Solution

a)

mean = 183

standard deviation = 20

z-score at 95% = 1.96

n= size of sample = 50

step1:
calculate standard error of mean

standard error of mean = z-score * standard deviation/sqrt(n)

= 1.96 * 20/sqrt(50)

= 5.544

step 2:

upper limit = mean + standard error

= 183 + 5.544

= 188.54

lower limit = mean - standard error

=183 - 5.544

= 177.46

hence

the interval is from 177.46 to 188.54

b)

mean =860

standard deviation = 12

z-score at 95% = 1.96

n= size of sample = 6

step1:
calculate standard error of mean

standard error of mean = z-score * standard deviation/sqrt(n)

= 1.96 * 12/sqrt(6)

= 9.60

step 2:

upper limit = mean + standard error

= 860 + 9.60

= 869.60

lower limit = mean - standard error

=860 -9.60

= 850.40

hence

the interval is from 850.40 to 869.60

C)

mean = 77

standard deviation = 4

z-score at 95% = 1.96

n= size of sample = 27

step1:
calculate standard error of mean

standard error of mean = z-score * standard deviation/sqrt(n)

= 1.96 * 4/sqrt(27)

= 1.509

step 2:

upper limit = mean + standard error

= 77 + 1.509

= 78.509

lower limit = mean - standard error

=77 - 1.509

= 75.491

hence

the interval is from 75.491 to 78.509