Question

The mean incubation time of fertilized eggs is 19 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day.

​(a) Determine the 10th percentile for incubation times.

​(b) Determine the incubation times that make up the middle 95​% of fertilized eggs.

Answer 1

solution

Using standard normal table,

P(Z < z) = 10%

= P(Z < z) = 0.10

= P(Z < -1.28) = 0.10

z = -1.28 Using standard normal z table,

Using z-score formula  

x= z * +

x= -1.28 *1+19

x= 17.72

(B)

middle % of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0. 95= 1.95

P(Z < z) = 1.95/ 2 = 0.975

P(Z <1.96 ) = 0.975

z  ±1.96   

Using z-score formula  

x= z * +

x= - 1.96*1+19

x= 17.04

z = 1.96

Using z-score formula  

x= z * +

x=1.96 *1+19

x= 20.96