you can use the formula for sample means:
[tex][tex]z(70) = \frac{70-75}{5 / \sqrt{64}} = -8[/tex][/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:
[tex]z(70) = \frac{70-75}{5 / \sqrt{64}} = -8[/tex]
[tex]z(80) = \frac{80-75}{5 / \sqrt{64}} = 8[/tex]
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
