The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Answer the two questions here : A) In a letter to an advice column, a wife claimed to have given birth 309 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 309 days or longer. What does the result suggest?

We are asked to find the probability of a pregnancy lasting **309 days or longer**, given that the lengths of pregnancies are **normally distributed** with a **mean** of 268 days and a **standard deviation** of 15 days. Here's the step-by-step solution:

### Step 1: Standardize the value using the Z-score formula
To find the probability of a pregnancy lasting 309 days or longer, we first convert the number of days (309 days) into a **Z-score**. The Z-score formula is:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \( X \) is the value (309 days),
- \( \mu \) is the mean (268 days),
- \( \sigma \) is the standard deviation (15 days).

Substituting the values into the formula:

\[
Z = \frac{309 - 268}{15} = \frac{41}{15} = 2.73
\]

### Step 2: Find the probability corresponding to the Z-score
Next, we need to find the probability that a Z-score is **greater than 2.73**. We can use standard normal distribution tables or a calculator to find this probability.

From the **Z-table** or a standard normal distribution calculator, the cumulative probability of a Z-score of 2.73 is approximately **0.9968**. This means the probability of a Z-score **less than** 2.73 is 99.68%.

To find the probability of a Z-score **greater than** 2.73, we subtract this value from 1:

\[
P(Z > 2.73) = 1 - 0.9968 = 0.0032
\]

### Step 3: Interpret the result
The probability of a pregnancy lasting 309 days or longer is **0.32%** (0.0032). This is an extremely small probability, which suggests that such a long pregnancy is **very rare**. 

### What does this suggest?
Given that the likelihood of a pregnancy lasting 309 days or longer is only 0.32%, the result suggests that it is **highly unlikely** that the wife gave birth 309 days after the husband's visit. Therefore, the claim might raise questions about the timeline she provided.

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Would you like further details or clarifications? Here are some related questions:

1. How do you interpret the standard deviation in this context?
2. What does the Z-score represent in normal distribution problems?
3. Could a different standard deviation change the outcome significantly?
4. What are other real-life situations where the Z-score is used?
5. How does this normal distribution apply to everyday occurrences?

**Tip**: In normal distributions, data values far from the mean (in terms of Z-scores) are considered outliers or rare events.

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Answer the two questions here: A) In a letter to an advice column, a wife claimed to have given birth 309 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 309 days or longer. What does the result suggest?

Calculate the probability of a pregnancy lasting 309 days or longer using the normal distribution. Given a mean of 268 days and a standard deviation of 15 days, this problem explores how to use the Z-score to find the likelihood of rare events. Suitable for Grades 10-12.

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