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? @brittneeharley

Answer 1

The first thing you want to do is find the z score, right?

Answer 2

Keep in mind we are looking at the sampling distribution of sample means . So we have to use standard error , standard deviation / sqrt( n )
$$ \Large z = \frac{ 70 - 75 }{ \frac{15 }{ \sqrt{64}}}$$

Answer 3

so thats the first z score

Answer 4

$$ \Large{
z = \frac{ 70 - 75 }{ \frac{15 }{ \sqrt{64}}}= -2.667
\\~\\
z = \frac{ 80 - 75 }{ \frac{15 }{ \sqrt{64}}}= 2.667
}
$$

Answer 5

so you need to find the area