he mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. (a) Determine the 17th percentile for incubation times. (b) Determine the incubation times that make up the middle 97%. LOADING... Click the icon to view a table of areas under the normal curve. (a) The 17th percentile for incubation times is nothing days. (Round to the nearest whole number as needed.) (b) The incubation times that make up the middle 97% are nothing to nothing days. (Round to the nearest whole number as needed. Use ascending order.)

Normal distribution is a curve plotted when a symmetrical data set is plotted on a graph with freqency on y-axis and data values on x-axis.

It is a bell shaped curve

What is z-score?

It is given by the formula:

[tex]z=\frac{|x-\mu|}{\sigma}[/tex]

where x is the value whose z score is to be computed, [tex]\mu[/tex] is is the population mean and [tex]\sigma[/tex] is the standard deviation.

a) To calculate centiles one must refer to the Z table

In the z table search for value just greater than .17

So for .17 the z-score is -0.95

Now the value can be computed by the formula

[tex]x=\mu+z\sigma[/tex]

As per question, [tex]\mu[/tex] = 23 and [tex]\sigma[/tex] = 1

x = 23 + (-0.95)*1

x= 22.05

So the The 17th percentile for incubation times is 22 days.

b) T calculate middle 97% we must find the values corresponding to 1.5% and 98.5%. Using the z table, we find that for 1.5% z = -2.97 and 98.5% z=+2.97

Therefore by using the same formula as in (a)

[tex]x=\mu+z\sigma[/tex]

x = 23 + (±2.97)*1

x= 20.03 and 25.97

So The incubation times that make up the middle 97% are 20 to 26 days.

earn more about z tables here https://brainly.com/question/7001627

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