In: Statistics and Probability
how do you get the answers using a calculator. the answers is 0.9940. my professor gave us a review with answers but didn't explain the answers. she wrote 2nd VARS (2) normal cdf ( lower, upper, u, sd)
18. In the past, all patrons of a cinema complex have spent an average of $5.00 for popcorn and other snacks, with a standard deviation of $1.80. If a random sample of 32 patrons is taken, what is the probability that the mean expenditure of this sample is greater than $4.20?
Solution :
Given that ,
mean =
= 5.00
standard deviation =
= 1.80
n = 32
=
= 5.00 and
=
/
n = 1.80 /
32 = 0.3182
P(
> 4.20) = 1 - P(
< 4.20)
= 1 - P((
-
) /
< (4.20 - 5.00) / 0.3182)
= 1 - P(z < -2.51)
= 1 - 0.0060
= 0.9940
Probability = 0.9940