Answer to Question #2373 in Statistics and Probability for bx
In a study for Health Statistics, it was found that the mean height of female 20-25 years old was 64.1 inches. If height is normally distributed with a standard deviation of 2.8 inches.
a) Determine the height required to be in the top 10% of all 20-25 years old females.
b) A “one-size-fits all” robe is being designed that should fit 90% of 20-25 years old females; what heights constitute the middle 90% of all 20-25 years old females?
Let X be height of female 20-25 years old, then Z = (X-64.1)/2.8 has standard normal distribution. a) We have to such H that P(X>H)=0.1 0.1 = P(X>H) = P( Z>(H-64.1)/2.8 ) = = 1 - F((H-64.1)/2.8) So F((H-64.1)/2.8) = 0.9 Then, using known values of F we obtain that F((H-64.1)/2.8)= 1.281552 Hence H =1.281552 * 2.8 + 64.1 = 67.69
b) We have to find D such that P(64.1-D < X < 64.1+D ) = 0.9