Question

The mean incubation time of fertilized eggs is 20 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. ​(a) Determine the 14th percentile for incubation times. ​(b) Determine the incubation times that make up the middle 95​% of fertilized eggs.

Answer 1

Part a)
X ~ N ( µ = 20 , σ = 1 )
P ( X < x ) = 14% = 0.14
To find the value of x
Looking for the probability 0.14 in standard normal table to calculate Z score = -1.0803
Z = ( X - µ ) / σ
-1.0803 = ( X - 20 ) / 1
X = 18.9197
P ( X < 18.9197 ) = 0.14

Part b)
X ~ N ( µ = 20 , σ = 1 )
P ( a < X < b ) = 0.95
Dividing the area 0.95 in two parts we get 0.95/2 = 0.475
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.475
Area above the mean is b = 0.5 + 0.475
Looking for the probability 0.025 in standard normal table to calculate Z score = -1.96
Looking for the probability 0.975 in standard normal table to calculate Z score = 1.96
Z = ( X - µ ) / σ
-1.96 = ( X - 20 ) / 1
a = 18.04
1.96 = ( X - 20 ) / 1
b = 21.96

P ( 18.04 < X < 21.96 ) = 0.95