Question

Assume that the number of advertisements that an individual sees in a given day follows a Normal distributionN(25,4). That is, the population mean is 25 advertisements, and the population standard deviation is 4 advertisements.

A. What is the likelihood that an individual sees more than 30 advertisements?

B. What is the likelihood that 9 individuals see more than an average of 30 advertisements?

C. What is the likelihood that 9 individuals see an average between 20-30 advertisements?

D. For a given individual, what is the 90^th percentile for the number of advertisements seen? (That is, how many advertisements would someone see if they saw more advertisements than 90% of all other individuals in the population?)

Answer 1

A. Here we need to find

As distribution is normal we can convert x to z

b. Now we need to find

As population is normal we can convert sample mean to z

c. Now we need to find

d. Here we need to find x such that

Using z table we get

So

So