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 20th percentile for incubation times. (b) Determine the incubation times that make up the middle 97% of fertilized eggs.

Step-by-step explanation:

Since 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 of fertilized eggs in days

µ = mean incubation time

σ = standard deviation

From the information given,

µ = 19 days

σ = 1 day

a) For the 20th percentile for incubation times, it means that 20% of the incubation times are below or even equal to 19 days(on the left side). We would determine the z score corresponding to 20%(20/100 = 0.2)

Looking at the normal distribution table, the z score corresponding to the probability value is - 0.84

Therefore,

- 0.84 = (x - 19)/1

x = - 0.84 + 19 = 18.16

b) for the incubation times that make up the middle 97% of fertilized eggs, the probability is 97% that the incubation times lie below and above 19 days. Thus, we would determine 2 z values. From the normal distribution table, the two z values corresponding to 0.97 are

1.89 and - 1.89

For z = 1.89,

1.89 = (x - 19)/1

x = 1.89 + 19 = 20.89 days

For z = - 1.89,

- 1.89 = (x - 19)/1

x = - 1.89 + 19 = 17.11 days

the incubation times that make up the middle 97% of fertilized eggs are

17.11 days and 20.89 days

take it another

d i think

step-by-step explanation: