Based on data from the National Center for Health Statistics, N. Wetzel used the normal distribution to model the length of gestation for pregnant US woman ( Chance, Spring 2001). Gestation length has mean of 280 days with a standard deviation of 20 days.

a. Find the probability that gestation length is between 275.5 and 276.5 days. (This estimates the probability that a woman has her baby 4 days earlier than the due date)
b. Find probability that gestation length is between 258.5 and 259.5 days. (This estimates the probability that a woman has her baby 21 days earlier than the due date).
c. Find the probability that gestation length is between 254.5 and 255.5 days. (This estimates the probability that a woman has her baby 25 days earlier than the due date.
d. The Chance article referenced a newspaper story about three sisters who all gave birth on the same day (March 11, 1998). Karralee had her baby 4 days early: Marrianne had her baby 21 days early: and Jennifer had her baby 25 days early. Use the results, parts a-c estimate the probability that all three woman have their babies 4, 21, 25 days early, respectively. Assume the births are independent events.

a-c. Z = (score-mean)/SD

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

d. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

To solve these probability questions, we will use the standard normal distribution, since we have the mean and standard deviation.

Let's denote the mean (μ) as 280 days and the standard deviation (σ) as 20 days.

a. To find the probability that gestation length is between 275.5 and 276.5 days (4 days earlier than the due date), we need to find the area under the normal distribution curve between these two values.

Let's calculate the z-scores for both values:
For 275.5 days: z = (275.5 - 280) / 20 = -0.225
For 276.5 days: z = (276.5 - 280) / 20 = -0.175

Now, we need to find the probability associated with these z-scores. We can do this using a standard normal distribution table or calculator.

Using a standard normal distribution table, we find that the probability of having a z-score between -0.225 and -0.175 (inclusive) is approximately 0.034.

So, the probability that gestation length is between 275.5 and 276.5 days is approximately 0.034.

b. To find the probability that gestation length is between 258.5 and 259.5 days (21 days earlier than the due date), we follow the same steps as in part a.

The z-score for 258.5 days is: z = (258.5 - 280) / 20 = -1.075
The z-score for 259.5 days is: z = (259.5 - 280) / 20 = -1.025

Using the standard normal distribution table, we find that the probability of having a z-score between -1.075 and -1.025 (inclusive) is approximately 0.070.

So, the probability that gestation length is between 258.5 and 259.5 days is approximately 0.070.

c. To find the probability that gestation length is between 254.5 and 255.5 days (25 days earlier than the due date), we follow the same steps as in parts a and b.

The z-score for 254.5 days is: z = (254.5 - 280) / 20 = -1.275
The z-score for 255.5 days is: z = (255.5 - 280) / 20 = -1.225

Using the standard normal distribution table, we find that the probability of having a z-score between -1.275 and -1.225 (inclusive) is approximately 0.048.

So, the probability that gestation length is between 254.5 and 255.5 days is approximately 0.048.

d. Since the births are assumed to be independent events, we can multiply the probabilities obtained in parts a, b, and c to find the probability that all three women have their babies 4, 21, and 25 days early, respectively.

The probability that all three women have their babies 4, 21, and 25 days early is approximately:
0.034 (from part a) * 0.070 (from part b) * 0.048 (from part c) ≈ 0.000118 or 0.0118%.

So, the estimated probability that all three women have their babies 4, 21, and 25 days early, respectively, is approximately 0.0118%.

To solve these problems, we will use the normal distribution formula to find the probability of gestation length falling within a specific range. The formula is:

P(a < X < b) = Φ((b - μ) / σ) - Φ((a - μ) / σ)

Where:
P(a < X < b) is the probability that X falls between a and b,
Φ(z) is the cumulative distribution function (CDF) of the standard normal distribution,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.

Now let's solve each problem step by step:

a. Find the probability that gestation length is between 275.5 and 276.5 days.
Here, a = 275.5, b = 276.5, μ = 280, and σ = 20.
Plug these values into the formula:

P(275.5 < X < 276.5) = Φ((276.5 - 280) / 20) - Φ((275.5 - 280) / 20)

Calculate the z-scores:

Z1 = (276.5 - 280) / 20 ≈ -0.175
Z2 = (275.5 - 280) / 20 ≈ -0.225

Look up the corresponding probabilities using a standard normal distribution table or a calculator. The result will be the probability of gestation length falling between 275.5 and 276.5 days.

b. Find the probability that gestation length is between 258.5 and 259.5 days.
Follow the same steps as in part a, with a = 258.5, b = 259.5, μ = 280, and σ = 20.

P(258.5 < X < 259.5) = Φ((259.5 - 280) / 20) - Φ((258.5 - 280) / 20)

Calculate the z-scores:

Z1 = (259.5 - 280) / 20 ≈ -1.025
Z2 = (258.5 - 280) / 20 ≈ -1.075

Look up the corresponding probabilities using a standard normal distribution table or a calculator.

c. Find the probability that gestation length is between 254.5 and 255.5 days.
Follow the same steps as in part a, with a = 254.5, b = 255.5, μ = 280, and σ = 20.

P(254.5 < X < 255.5) = Φ((255.5 - 280) / 20) - Φ((254.5 - 280) / 20)

Calculate the z-scores:

Z1 = (255.5 - 280) / 20 ≈ -1.225
Z2 = (254.5 - 280) / 20 ≈ -1.275

Look up the corresponding probabilities using a standard normal distribution table or a calculator.

d. Estimate the probability that all three women have their babies 4, 21, and 25 days early, respectively.
Since the births are assumed to be independent events, we can multiply the probabilities from parts a, b, and c to find the probability of all three events happening:

P(all three events) = P(275.5 < X < 276.5) * P(258.5 < X < 259.5) * P(254.5 < X < 255.5)

Multiply the probabilities obtained in parts a, b, and c, respectively, to find the desired probability.

Note: When looking up probabilities in a standard normal distribution table, you typically find the value for the cumulative probability up to a specific z-score. Subtracting one cumulative probability from another gives you the probability between two z-scores.