Tompkins Associates reports that the mean clear height for a Class A warehouse in the United Statesis 22 feet. Suppose clear heights are normally distributed and that the standard deviation is 4 feet. A Class A warehouse in the United States is randomly selected.A) What is the probability that the clear height is greater than 18 feet? P(x>18) = 0.8413B) What is the probability that the clear height is less than 11 feet? P(x<11) = ???C) What is the probability that the clear height is between 23 and 31 feet? P(23<= x <= 31) = ???(Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

Answer:A) 84.13%B) 0.3%C) 38.91%Step-by-step explanation:We standardize the values of x by using the formula  [tex]\bf z=\frac{x-\mu}{\sigma}[/tex] where  [tex]\bf \mu[/tex] is the mean = 22 feet [tex]\bf \sigma[/tex] is the standard deviation = 4 feet By doing this, we transform the original distribution too the normal distribution N(0;1) and we can compute the probabilities easily by looking up at tables or by using the computer A) for x = 18  [tex]\bf z=\frac{18-22}{4}=-1[/tex] So P(x > 18) = P(z > -1) = 0.8413 or 84.13% B) In this case, since a height cannot be negative, we really want P(0 < x < 11) for x= 0 and x = 11, z = -5 and z = -2.75 and P(0 < x < 11) = P(-5 < z < -2.75) = 0.003 = 0.3% C) For x = 23 and x = 31, z = 0.25 and z = 2.25 and [tex]\bf P(23\leq x \leq 31) = P(0.25\leq z \leq 2.25) = 0.3891=38.91\%[/tex]