Question

A day care program has an average daily expense of $75.00. The standard deviation is $15.00. The owner takes a sample of 64 bills. What is the probability the mean of his sample will be between $70.00 and $80.00?

Answer 1

Central theorem.

Answer 2

mean \(\mu =75\), standard deviation = 15, n =64
\(\bar X\) is sample mean
\(P(70< \bar X <80)\) = ?
Apply Central theorem
\(P(\dfrac{70-\mu}{\sigma/\sqrt n}<\dfrac{\bar X-\mu }{\sigma/\sqrt n}<\dfrac{80-\mu}{\sigma/\sqrt n})=?\)

Answer 3

the middle term is normal distribution \(N\sim (75,15)\)
you just put all in calculator.