The mean incubation time of fertilized eggs is 21 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day.(a) Determine the 20th percentile for incubation times.(b) Determine the incubation times that make up the middle 39% of fertilized eggs.

Answer:a) 20.16; b) 20.49 and 21.51Step-by-step explanation:We use z scores for each of these.  The formula for a z score is[tex]z=\frac{X-\mu}{\sigma}[/tex].For part a, we want the 20th percentile; this means we want 20% of the data to be lower than this. We find the value in the cells of the z table that are the closest to 0.20 as we can get; this is 0.2005, which corresponds with a z score of -0.84.Using this, 21 as the mean and 1 as the standard deviation,-0.84 = (X-21)/1-0.84 = X-21Add 21 to each side:-0.84+21 = X-21+2120.16 = XFor part b, we want the middle 39%.  This means we want 39/2 = 19.5% above the mean and 19.5% below the mean; this means we want50-19.5 = 30.5% = 0.305 and50+19.5 = 69.5% = 0.695.Looking these values up in the cells of the z table, we find those exact values; 0.305 corresponds with z = -0.51 and 0.695 corresponds with z = 0.51:-0.51 = (X-21)/1-0.51 = X-21Add 21 to each side:-0.51+21 = X-21+2120.49 = X0.51 = (X-21)/10.51 = X-21Add 21 to each side:0.51+21 = X-21+2121.51 = X