The incubation period for the eggs of a house wren is normally distributed with a mean of 336 hours and a standard deviation of 3.5 hours. Find the probability that an egg has an incubation period between 328 and 338 hours.
Suppose a researcher has fifteen house wren eggs in an incubator. Find the probability that the average incubation time for the fifteen eggs is greater than 337 hours.
Suppose it is known that 33% of House wren eggs never hatch. Suppose an ornithologist has forty house wren eggs. Find the probability that less than 25% never hatch.

a) 0.705015

b) 0.13424

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

The incubation period for the eggs of a house wren is normally distributed with a mean of 336 hours and a standard deviation of 3.5 hours.

a) Find the probability that an egg has an incubation period between 328 and 338 hours.

For x = 328

z = 328 - 336/3.5

z =-2.28571

Probability value from Z-Table:

P(x = 328) = 0.011135

For x = 338

z = 338 - 336/3.5

z =0.57143

Probability value from Z-Table:

P(x = 338) = 0.71615

The probability that an egg has an incubation period between 328 and 338 hours.

P(x = 338) - (P(x = 328)

0.71615 - 0.011135

= 0.705015

b) Suppose a researcher has fifteen house wren eggs in an incubator. Find the probability that the average incubation time for the fifteen eggs is greater than 337 hours.

We are given a sample of 15 wren eggs,

We use this z score formula

z = (x-μ)/σ/√n, where

x is the raw score

μ is the population mean

σ is the population standard deviation

n is random number of samples

z = 337 - 336/3.5/√15

z =1.10657

Probability value from Z-Table:

P(x<337) = 0.86576

P(x>337) = 1 - P(x<337) = 0.13424

Suppose it is known that 33% of House wren eggs never hatch. Suppose an ornithologist has forty house wren eggs. Find the probability that less than 25% never hatch.