Question

Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with mean 219.5 ft and standard deviation 62.5 ft You intend to measure a random sample of n 187 trees. The bell curve below represents the distibution of these sample means. The scale on the horizontal axis is the standard error of the sampling distribution Complete the indicated boxes, correct to two decimal places Points possible: 1 Unlimited attempts 8:11 PM 1/22/2019 EWLETTPACKARD BANG OLOFSE prt sc 5 6 8 9 +back
Please help ASAP please.

Answer 1

Solution :- Given that mean = 219.5, standard deviation = 62.5 , n = 187

μx = 219.5 σx = 62.5/sqrt(187) = 4.57

=> Lower limit : 219.5 - (2*4.57) = 210.36

=> middle : 219.5

=> upper limit : 219.5 + (2*4.57) = 228.64

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Solution:-

That is we need sample proportion between 0.75-0.03 = 0.72 and 0.75+0.03 = 0.78.
Here we have p=0.75 and n=159. So z-score for sample proporiton p^ =0.72 is

Z = (p^-p)/sqrt((pq/n)) = (0.72-0.75)/sqrt(0.75*0.25/159) = -0.8736

z-score for sample proporiton p^ = 0.78

Z = (p^-p)/sqrt((pq/n)) = (0.78-0.75)/sqrt(0.75*0.25/159) = 0.8736

=> P(-0.8736 < Z < 0.8736) = 0.6156

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