Question

The mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. Find the incubation time that separates the bottom 2.5​% from the rest of the incubation times.

Answer 1

Answer: the incubation time that separates the bottom 2.5​% from the rest of the incubation times is 24.96 days.

Step-by-step explanation:

Suppose 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.

µ = mean time

σ = standard deviation

From the information given,

µ = 23 days

σ = 1 day

The probability value for the incubation time that separates the bottom 2.5​% from the rest of the incubation times would be (1 - 2.5/100) = (1 - 0.025) = 0.975

Looking at the normal distribution table, the z score corresponding to the probability value is 1.96

Therefore,

1.96 = (x - 23)/1

x = 1.96 + 23

x = 24.96

Answer 2

Answer:

The incubation time that separates the bottom 2.5​% from the rest of the incubation times is 21.04 days.

Step-by-step explanation:

We are given that the mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day.

Let X = the incubation times

So, X ~ N($\mu=23,\sigma^{2} = 1^{2}$)

The z score probability distribution is given by;

         Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean incubation time = 23 days

            $\sigma$ = standard deviation = 1 day

Now, we have to find the incubation time that separates the bottom 2.5​% from the rest of the incubation times.

So, Probability that the incubation time separate the bottom 2.5​% is given by;

        P(X > x) = 0.025

        P( $\frac{X-\mu}{\sigma}$ > $\frac{x-23}{1}$ ) = 0.025

        P(Z > $\frac{x-23}{1}$ ) = 0.025

So, the critical value of x in z table which separate the bottom 2.5% is given as -1.96, which means;

                  $\frac{x-23}{1}$  = -1.96

                        $x$ = 23 - 1.96 = 21.04

Therefore, the incubation time that separates the bottom 2.5​% from the rest of the incubation times is 21.04 days.