Question 1161685: Americans ate an average of 25.7 pounds of confectionary 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 Boreal(15235)   (Show Source):
You can put this solution on YOUR website! this is a z-test
z=(xbar-mean)/sigma/sqrt(n)
=(27-25.7)/3.75/sqrt(40)
=1.3*sqrt(40)/3.75
=2.19
it is the probability z>2.19, which is 0.0143
calculator 2nd VARS 2 normalcdf ENTER (27,1000,25.7,3.75/sqrt(40) ENTER. The 1000 is just a large number on the infinite upside. You can use 1E99 if desired, That has less rounding and gives 0.0142.
z=(60-61.50)/5.89/sqrt(50)
>-1.5*sqrt(50)/5.89
>-1.80 or probability 0.9641
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