To solve these problems, we will use properties of the normal distribution. The standard normal distribution $Z$ has a mean of 0 and a standard deviation of 1, and we can convert any normally distributed variable $X$ with mean $\mu$ and standard deviation $\sigma$ into the standard normal variable $Z$ using the formula:

$Z = \frac{X - \mu}{\sigma}$

Given:

• Mean ($\mu$) = 266 days
• Standard deviation ($\sigma$) = 16 days

Let's address each question:

1. Percentage of pregnancies lasting between 234 and 298 days

We need to find the percentage of pregnancies that last between 234 and 298 days. First, we convert these values to their corresponding Z-scores:

$Z_1 = \frac{234 - 266}{16} = \frac{-32}{16} = -2$ $Z_2 = \frac{298 - 266}{16} = \frac{32}{16} = 2$

Next, we find the probability corresponding to these Z-scores using the standard normal distribution table or a calculator:

$P(Z < 2) - P(Z < -2)$

Using the Z-table:

$P(Z < 2) \approx 0.9772 \quad \text{and} \quad P(Z < -2) \approx 0.0228$

So, the percentage of pregnancies lasting between 234 and 298 days is:

$0.9772 - 0.0228 = 0.9544 \quad \text{or} \quad 95.44\%$

2. Percentage of gestation times exceeding 312 days

We need to find the percentage of pregnancies that last more than 312 days. First, convert 312 days to its Z-score:

$Z = \frac{312 - 266}{16} = \frac{46}{16} = 2.875$

Next, find the probability:

$P(Z > 2.875)$

Using the Z-table:

$P(Z < 2.875) \approx 0.9979$

So, the percentage exceeding 312 days is:

$1 - 0.9979 = 0.0021 \quad \text{or} \quad 0.21\%$

3. Probability that at least one woman out of 100 has a gestation period of more than 312 days

The probability that one woman has a gestation period of more than 312 days is 0.0021, as calculated earlier. The probability that a woman does not have a gestation period exceeding 312 days is:

$P(\text{Not exceed}) = 1 - 0.0021 = 0.9979$

The probability that all 100 women have a gestation period of 312 days or less is:

$P(\text{All 100 not exceed}) = 0.9979^{100} \approx 0.7884$

Thus, the probability that at least one of the 100 women has a gestation period of more than 312 days is:

$1 - 0.7884 = 0.2116 \quad \text{or} \quad 21.16\%$

4. Length of gestation exceeded by only 5% of pregnancies

We need to find the gestation time $X$ such that only 5% of pregnancies exceed this time. This corresponds to finding the Z-score for the 95th percentile:

$P(Z < Z_{\text{95\%}}) = 0.95$

From the Z-table:

$Z_{\text{95\%}} \approx 1.645$

Now, convert this Z-score back to the original gestation time $X$:

$X = \mu + Z \cdot \sigma = 266 + 1.645 \cdot 16 = 266 + 26.32 \approx 292 \, \text{days}$

So, only 5% of pregnancies exceed 292 days.

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Would you like any more details or have any questions? Here are five related questions you might be interested in:

1. What percentage of pregnancies last between 250 and 280 days?
2. What is the probability that a pregnancy lasts less than 230 days?
3. If a pregnancy lasts 280 days, what is the Z-score associated with this length?
4. What is the probability that two randomly chosen women both have pregnancies exceeding 300 days?
5. How would the percentage of pregnancies exceeding 312 days change if the standard deviation were 10 days instead of 16?

Tip: Understanding how to use the Z-table is crucial for solving problems involving the normal distribution. It allows you to find probabilities for any normally distributed variable quickly.