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 14th percentile for incubation times.
(b) determine the incubation times that make up the middle 97% of fertilized eggs.
(a) the 14th percentile for incubation times is 18 days.
(round to the nearest whole number as needed.)
(b) the incubation times that make up the middle 97% of fertilized eggs are \\( \square \\) to \\( \square \\) days.
(round to the nearest whole number as needed. use ascending order.)

Explanation:

Step1: Find the z - score for the middle 97%

The middle 97% leaves \(100\% - 97\%=3\%\) in the two tails. So, \( \alpha=\frac{3\%}{2}=0.015\) in each tail.
Looking up the z - score in the standard normal table, \(z_{0.015}\) is such that \(P(Z < z_{0.015})=0.015\) and \(P(Z>z_{1 - 0.015}) = 0.015\). The z - score \(z_{0.015}\approx - 2.17\) and \(z_{0.985}\approx2.17\) (using standard normal distribution tables or a calculator with a normal - distribution function).

Step2: Use the z - score formula \(x=\mu+z\sigma\)

Given \(\mu = 19\) (mean) and \(\sigma = 1\) (standard deviation).
For \(z=-2.17\):
\(x_1=\mu+z\sigma=19+(- 2.17)\times1=19 - 2.17 = 16.83\)
For \(z = 2.17\):
\(x_2=\mu+z\sigma=19+2.17\times1=19 + 2.17=21.17\)

Answer:

The incubation times that make up the middle 97% of fertilized eggs are \(17\) to \(21\) days.