SOLUTION: Americans ate an average of
25.7 pounds of confectionery products each last year
and spent an average of $61.50 per person doing so. If
the standard deviation for consumption is
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-> SOLUTION: Americans ate an average of
25.7 pounds of confectionery products each last year
and spent an average of $61.50 per person doing so. If
the standard deviation for consumption is
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Question 1161604: Americans ate an average of
25.7 pounds of confectionery products each last year
and spent an average of $61.50 per person doing so. If
the standard deviation for consumption is 3.75 pounds
and the standard deviation for the amount spent is
$5.89, find the following:
a. The probability that the sample mean confectionary
consumption for a random sample of 40 American
consumers was greater than 27 pounds
b. The probability that for a random sample of 50, the
sample mean for confectionary spending exceeded
$60.00 Answer by Theo(13342) (Show Source):
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 http://davidmlane.com/hyperstat/z_table.html, 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:
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 http://davidmlane.com/hyperstat/z_table.html, 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:
