In: Math
6.20. A convenience store owner wants to know how long his customers spend browsing the store before making a purchase. It is found that time spent is normally distributed with an average of m = 5 minutes and a standard deviation of s=2 : 2 minutes. Using a random sample of 14 customers, what is the probability that a customer, on average, will spend less than 4 minutes browsing the store?
Solution :
Given that ,
mean =
= 5
standard deviation =
= 2
n = 14
=
= 5 and
=
/
n = 2/
14 = 0.5345
P(
< 4) = P((
-
) /
< (4 - 5) / 0.5345)
= P(z < -1.87)
= 0.0307 Using standard normal table,
The probability that a customer, on average, will spend less than 4 minutes browsing the store is 0.0307.