The mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. Find the incubation time that separates the bottom 2.5% from the rest of the incubation times.

the incubation time that separates the bottom 2.5% from the rest of the incubation times is 24.96 days.

Step-by-step explanation:

Suppose the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = incubation times.

µ = mean time

σ = standard deviation

From the information given,

µ = 23 days

σ = 1 day

The probability value for the incubation time that separates the bottom 2.5% from the rest of the incubation times would be (1 - 2.5/100) = (1 - 0.025) = 0.975

Looking at the normal distribution table, the z score corresponding to the probability value is 1.96

Therefore,

1.96 = (x - 23)/1

x = 1.96 + 23

x = 24.96