Q:

The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.9 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to 4 decimal places where possible.a. What is the distribution of X? X ~ N([],[])b. What is the median giraffe height? [] ft.c. What is the Z-score for a giraffe that is 18.5 foot tall? []d. What is the probability that a randomly selected giraffe will be shorter than 18 feet tall?[] e. What is the probability that a randomly selected giraffe will be between 16.9 and 17.9 feet tall? []f. The 75th percentile for the height of giraffes is [] ft.

Accepted Solution

A:
Answer:Median =17Step-by-step explanation:Given that the mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.9 feet. Let X be the height of a randomly selected adult giraffe.X is a random variable. Β a) X is N(17, 0.9)b) Median giraffe height = 17 ft since in normal distribution mean = medianc) When x = 18.5, Z = [tex]\frac{18.5-17}{0.9} =0.1667[/tex]d) [tex]P(X<18) = P(Z<0.56)\\=0.2877[/tex]e) [tex]P(16.9<x<17.9) = P(-0.11<z<1) = 0.0438+0.3413\\=0.3851[/tex]f) 75th percentile for z is 0.675[tex]x=17.9+0.9(0.675)\\= 18.5075[/tex]