Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 269 days and standard deviation sigma equals 22 days. Complete parts a. through c.

a. What is the probability that a randomly selected pregnancy lasts less than 262 days?
b. What is the probability that a random sample of 28 pregnancies has a mean gestation period of fewer than 262 days?
c. What is the probability that a random sample of 53 pregnancies has a mean gestation period of fewer than 262 days?

Answer 1

Answer:

a) 0.3745

b) 0.0465

c) 0.0102

Step-by-step explanation:

Lets call X the random variable that estimates the lenght of pregnancy (in days). By hypothesis X ≈ N(μ = 269,σ = 22)

a) First, we have to standarize X, we take $W = \frac{X - \mu}{\sigma} = \frac{X-269}{22}$ . W has Normal distribution with mean 0 and standard deviation equal to 1. The values of the cummulative function of W, Ф, are tabulated, and they can be obtained from the attached file. We want to calcualte P(X<262), we have

$P(X < 262) = P(\frac{X-269}{22} < \frac{262-269}{22}) = P(W < \frac{-7}{22}) = Φ(-0.318)$

We can use that the densitive function of a standard normal random variable is symetryc, therefore Φ(-0.318)  = 1- Φ(0.318)  = 1- 0.6255 = 0.3745

Thus, the probability for a pregnancy to last less than 262 days is 0.3745.

b) We call Z the mean gestation of 28 pregnancies. Z is the average of 28  independent random variables with distribution N(268,22).

The average of normal distributed random variables with identical distribution is also a normal random variable with similar mean, and the standard deviation is divided by √n, where n is the sample size. In this case, Z ≈ N(269,22/√28).

In order to calculate P(Z < 262), we standarize Z, obtaining a random variable $W = \frac{Z-269}{22/ \sqrt{28}} .$

$P(Z < 262) = P(W < \frac{262-269}{22 / \sqrt{28}}) = \phi(-1.68)$

Using the symetry of the density of W, we have that $\phi(-1.68) = 1-\phi(1.68) = 1-0.9535 = 0.0465$ .

Thus, the probability for the mean to be less than 262 is 0.0465

c) If the total size of the sample is 53 and we call the average lenght Z again, then Z ≈ N(μ = 269, σ=22/√53). The standarization of Z now is $W = \frac{Z-269}{22/\sqrt{53}} .$ Therefore

$P(Z < 262) = P(W < \frac{262-269}{22/\sqrt{53}}) = \phi(-2.316) = 1- \phi(2.316) = 1-0.9898 = 0.0102$

We conclude that the probability that the mean is less that 262 is 0.0102. Note that, the bigger the sample, the less likely is the average value from being 'far' from the mean.

I hope that works for you!