In: Math
You own a small company. Last year you conducted a study to learn more about your customers. You found that the mean age of your customers was 31.84 years with a standard deviation of 9.84 years. This year you take a random sample of 60 customers. What is the probability that the mean age of those 60 customers is greater than 33 years?
Solution :
Given that ,
mean =
= 31.84
standard deviation =
= 9.84
n = 60
= 31.84
=
/
n = 9.84 /
60 =1.2703
P(
> 33) = 1 - P(
< 33)
= 1 - P[(
-
) /
< (33 - 31.84) / 1.2703 ]
= 1 - P(z <0.91 )
Using z table,
= 1 - 0.8186
= 0.1814