Question 1161695: The average individual monthly spending in the United States for paging and messaging services is $10.15. If the standard deviation is $2.45 and the amounts are normally distributed, what is the probability that a randomly selected user of these services pays more than $15.00 per month? Between $12.00 and $14.00 per month?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
for >$15 a month
z>(15-10.15)/2.45 or 4.85/2.45 or 1.98
This has probability of 0.0239
between 12 and 14 would be done the same way, finding the z for each and looking for the z in between.
With the calculator, one can do it by 2nd VARS 2 normalcdf ENTER (12,14,10.15,2.45)=0.1671
for 12 z=+1.85/2.45 or +0.76
for 14 it would be 3.85/2.45 or +1.57
This would give an answer of 0.1654
with rounding of the z-values to 4 places, the answer is 0.1670
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