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 □ days.
(round to the nearest whole number as needed.)

Explanation:

Step1: Find the z - score for the 14th percentile

We know that if \(X\sim N(\mu,\sigma^{2})\), and we want to find the \(p\)th percentile. For the 14th percentile (\(p = 0.14\)), we look up the \(z\) - value in the standard normal distribution table (or use a calculator with a normal - distribution function). Using a standard normal table or a calculator (e.g., invNorm function in TI - 84: invNorm\((0.14)\)), the \(z\) - score \(z\approx - 1.08\).

Step2: Use the z - score formula to find \(x\)

The \(z\) - score formula is \(z=\frac{x-\mu}{\sigma}\). We are given that \(\mu = 19\) (mean) and \(\sigma = 1\) (standard deviation). Rearranging the formula for \(x\) gives \(x=\mu+z\sigma\).
Substitute \(\mu = 19\), \(z=-1.08\), and \(\sigma = 1\) into the formula: \(x=19+(-1.08)\times1\).

$$x=19 - 1.08=17.92$$

Answer:

\(18\)