Question 1161601: 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
z=(15-10.15)/2.45=1.978
probability z> 1.978 is 0.0239
between 12 and 14
do it the same way or on calculator 2nd VARS2 normal cdf ENTER (12,14,10.15,2.45)ENTER
0.1671
it is probability z is between 0.755 and 1.571 (0.1670), the difference in the last decimal place was that I was rounding in the second.
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