Question

To work as a flight attendant, you must be between 5ft 6 in and 6ft tall. The height of 18-24 year-old females is normally distributed with a mean of about 67.1 inches and a standard deviation of 2.4 inches.
[Hint:Convert feet into inches (1ft = 12 in)]


a) What is the percent of women ages 18-24 who are between 5ft 6 in and 6ft? %


b)What height represents 86 th percentile?


c)What is the probability that a female who is 18-24 years old is taller than 5ft 3 in? Round this to 4 decimal places

Answer 1

Solution:
Mean = 67.1
Standard deviation = 2.4
Solution(a)
We need to calculate
P(66<X<72) = P(X<72) - P(X<66)
Z = (72-67.1)/2.4 = 2.04
Z = (66-67.1)/2.4 = -0.46
From Z table we found p-value
P(66<X<72) = P(X<72) - P(X<66) = 0.9793 - 0.3228 = 0.6564
Solution(b)
p-value = 0.86, from Z table we found Z score = 1.08
1.08 = (X-67.1)/2.4
X = 69.692
Solution(c)
P(X>63)= 1 - P(X<=63)
Z = (63-67.1)/2.4 = -1.71
So from Z table we found p-value
P(X>63)= 1 - P(X<=63) = 1- 0.04363 = 0.9564