From past experience it is felt that the random variable, X, the age of a mother at the birth of her first child, is normally distributed with a mean of 21.3 and a variance of 16. Find the probability that a randomly selected mother:

(a) has her first child before age 16
(b) has her first child after age 30
(c) has her first child between the ages of 15 and 20
(d) has her first child between the ages of 14 and 28

Z = (score-mean)/SD

variance = SD^2

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

To find the probability in each of the given scenarios, we need to use the standard normal distribution and the Z-score.

The Z-score is a measure of how many standard deviations an observation or value is from the mean of a distribution. In this case, we can calculate the Z-score using the formula Z = (X - μ) / σ, where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

For each scenario, we will calculate the corresponding Z-scores and then use the standard normal distribution table to find the probabilities.

(a) To find the probability that a randomly selected mother has her first child before age 16, we need to find the Z-score for X = 16.

Z = (16 - 21.3) / √16 = -1.6875

Using the standard normal distribution table, we can find the area (probability) corresponding to a Z-score of -1.6875. This gives us the probability P(X < 16).

(b) To find the probability that a randomly selected mother has her first child after age 30, we need to find the Z-score for X = 30.

Z = (30 - 21.3) / √16 = 2.1687

Using the standard normal distribution table, we can find the area (probability) corresponding to a Z-score of 2.1687. This gives us the probability P(X > 30).

(c) To find the probability that a randomly selected mother has her first child between the ages of 15 and 20, we need to find the Z-scores for X = 15 and X = 20.

For X = 15:
Z = (15 - 21.3) / √16 = -1.9688

For X = 20:
Z = (20 - 21.3) / √16 = -0.8125

Using the standard normal distribution table, we can find the area (probability) corresponding to a Z-score of -1.9688 and -0.8125. Subtracting the probability corresponding to -1.9688 from the probability corresponding to -0.8125 gives us the desired probability P(15 < X < 20).

(d) To find the probability that a randomly selected mother has her first child between the ages of 14 and 28, we need to find the Z-scores for X = 14 and X = 28.

For X = 14:
Z = (14 - 21.3) / √16 = -1.75

For X = 28:
Z = (28 - 21.3) / √16 = 1.7812

Using the standard normal distribution table, we can find the area (probability) corresponding to a Z-score of -1.75 and 1.7812. Subtracting the probability corresponding to -1.75 from the probability corresponding to 1.7812 gives us the desired probability P(14 < X < 28).

Please note that the standard normal distribution table provides the cumulative probabilities (area under the curve) up to a given Z-score value. To find the desired probability, you can subtract the appropriate values from the table or use statistical software/calculators that provide the ability to calculate cumulative probabilities directly.