In the 2000 census, the so-called "long form'
reccived by onc of every six households contained
52 questions, ranging from your ocupation and
income all the way to whether you had a bathtub
According to the U.S. Census Bureau, the mean
completion time for the long form is 38
minutes.
Assuming a standard deviation of 5 minutes and a
simple random sample of 50 persons who flled out
the long form, what is the probability that their
average time for completion of the form was more than
45 minutes?

This problem involves the **sampling distribution of the sample mean**. We need to compute the probability that the sample mean completion time is more than 45 minutes. 

Since the population mean \(\mu = 38\) minutes and the population standard deviation \(\sigma = 5\) minutes are given, we can use the **central limit theorem (CLT)**. For a sufficiently large sample (in this case \(n = 50\)), the sampling distribution of the sample mean follows a normal distribution with the following characteristics:

- **Mean of the sampling distribution**: \(\mu_{\bar{X}} = \mu = 38\)
- **Standard error (SE) of the mean**:  
  \[
  \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{50}} = \frac{5}{7.071} \approx 0.707
  \]

---

### Step 1: Convert 45 minutes to a z-score
We want to calculate the **probability that the sample mean is more than 45 minutes**. For that, we convert the value of 45 minutes into a **z-score**:

\[
z = \frac{X - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{45 - 38}{0.707} = \frac{7}{0.707} \approx 9.9
\]

---

### Step 2: Find the probability using the z-table
A **z-score of 9.9** is extremely high. The area to the left of this z-score (which corresponds to the cumulative probability) is effectively 1. That means the probability that the sample mean is **greater than 45 minutes** is:

\[
P(\bar{X} > 45) = 1 - P(Z < 9.9) \approx 1 - 1 = 0
\]

---

### Final Answer
The probability that the average completion time for the sample of 50 people is **more than 45 minutes** is approximately:

\[
P(\bar{X} > 45) \approx 0
\]

This result makes sense since 45 minutes is far above the population mean (38 minutes), and with a small standard error of 0.707, it's extremely unlikely that a random sample of 50 people would take that long on average.

---

Would you like a deeper explanation or have any follow-up questions?  
Here are some related questions you might find helpful:

1. What is the central limit theorem and why is it important?
2. How do you calculate the standard error for different sample sizes?
3. What is the purpose of a z-score in hypothesis testing?
4. How can we use a normal distribution table for probability calculations?
5. How does increasing the sample size affect the standard error and the probability?

**Tip:** In practice, probabilities that yield extremely high z-scores (like 9.9) are treated as 0, because they are beyond the range of meaningful outcomes.

In the 2000 census, the so-called 'long form' received by one of every six households contained 52 questions, ranging from your occupation and income all the way to whether you had a bathtub. According to the U.S. Census Bureau, the mean completion time for the long form is 38 minutes. Assuming a standard deviation of 5 minutes and a simple random sample of 50 persons who filled out the long form, what is the probability that their average time for completion of the form was more than 45 minutes?

Learn how to calculate the probability that the average completion time for a sample of 50 people exceeds 45 minutes. This problem applies the Central Limit Theorem, sampling distribution, and z-scores using a population mean of 38 minutes and a standard deviation of 5 minutes. Suitable for college-level statistics students.

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