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.

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: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: