Question 1161604
question a.


find the probability that the sample mean confectionery
consumption for a random sample of 40 American
consumers was greater than 27 pounds


population mean = 25.7
population standard deviation = 3.75
sample size = 40
standard error = population standard deviation divided by square root of sample size = s = 3.75 / sqrt(40) = .59293 rounded to 5 decimal digits.
using the online statistics calculator found at <a href = "http://davidmlane.com/hyperstat/z_table.html" target = "_blank">http://davidmlane.com/hyperstat/z_table.html</a>, do the following:
select area from a value
set mean = 25.7
set standard deviation = .59293
set above = 27 and hit enter.
calculator tells you that the probability that the mean of a sample of size 40 being greater than 27 is equal to .0142.
here's what the results look like:
<img src = "http://theo.x10hosting.com/2020/081301.jpg" >


question b.


find the probability that, for a random sample of 50, the
sample mean for confectionery spending exceeded
$60.00


population mean = 61.5
population standard deviation = 5.89
sample size = 50
standard error = population standard deviation divided by square root of sample size = s = 5.89 / sqrt(50) = /.83297 rounded to 5 decimal places.
using the online statistics calculator found at <a href = "http://davidmlane.com/hyperstat/z_table.html" target = "_blank">http://davidmlane.com/hyperstat/z_table.html</a>, do the following:
select area from a value
set mean = 61.5
set standard deviation = .83297
set above = 60 and hit enter.
calculator tells you that the probability that the the mean of a sample of size 50 being greater than 60 is equal to 0.9641.
here's what the results look like:
<img src = "http://theo.x10hosting.com/2020/081302.jpg" >