Answer:
Step-by-step explanation:
It is given that Suppose that a travel bureau claims that the trees in a forest are 85 feet tall on average, with a standard deviation of 0.2 feet.
Here, Mean=85, Standard deviation=0.2, Sample=64.
Now, using the confidence interval as 95%, then z=1.96.
Thus, Margin error is given as:
[tex]ME=\frac{1.96{\times}0.2}{\sqrt{64}}=\frac{0.392}{8}[/tex]
[tex]ME=0.049[/tex]
Therefore, confidence interval: [tex]85-0.048<{\mu}<85+0.049[/tex]
that is: [tex]84.95<{\mu}<85.049[/tex]
Mean heights outside the confidence interval will be:
[tex]x\leq84.95[/tex] and [tex]x\geq85.049[/tex].
