Question

a day care program has an average daily expense of $75 the standard deviation is $5 the owner takes a sample of 64 bills what is the probability the mean of his sample will be between $70 and $80? The answer choices are 68 1.0 34 + 30. Step 1 calculate a z-score for $70 step to give the probability for Step 1. Step 3 calculate the Z score for $80. Step for give the probability for step 3.

Answer 1

you can use the formula for sample means:

$[tex]z(70) = \frac{70-75}{5 / \sqrt{64}} = -8$[/tex]

where m stands for a value of the sample mean.

We are looking for the value, specifically for its two borderline values: 70 and 80

and their cumulative probabilities, p(z(80)) and p(z(70)). The difference p(z(80))-p(z(70))

will give the probability that m falls in the interval [70,80]

So let's get cracking at it:

$z(70) = \frac{70-75}{5 / \sqrt{64}} = -8$

$z(80) = \frac{80-75}{5 / \sqrt{64}} = 8$

These are very large values of z. You may notice that any z-table online won't even bother covering range that high - the probabilities for these values are virtually 0 (in the neg case) 1 (in the pos case).

This means that numerically the probability of the sample mean of 64 samples falling within the range of 1 standard deviation is very close to 1

So the answer choice  should be 1.0