The amount of time travellers at an airport spend with customs officers has a mean of  μ =32 μ =32 seconds and a standard deviation of  σ =11 seconds. For a random sample of 45 travellers, what is the probability that their mean time spent with customs officers will be:

Standard Normal Distribution Table

**a.** Over 30 seconds?

Round to four decimal places if necessary

**b.** Under 35 seconds?

Round to four decimal places if necessary

**c.** Under 30 seconds or over 35 seconds?

Round to four decimal places if necessary

To solve these problems involving probabilities for sample means using the **Central Limit Theorem**, we proceed step by step.

Given:
- Population mean \( \mu = 32 \) seconds
- Population standard deviation \( \sigma = 11 \) seconds
- Sample size \( n = 45 \)

First, we calculate the **standard error** (SE) of the sample mean:

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{11}{\sqrt{45}} \approx 1.639
\]

### **a. Probability that the mean time is over 30 seconds**

We first calculate the **z-score** corresponding to a mean time of 30 seconds. The z-score is given by:

\[
z = \frac{X - \mu}{SE} = \frac{30 - 32}{1.639} \approx -1.221
\]

Next, we look up the z-score in the **standard normal distribution table** or use a calculator to find the corresponding probability.

For \( z = -1.221 \), the cumulative probability (area to the left of z) is approximately 0.1119. To find the probability that the mean time is **over** 30 seconds, we subtract this from 1:

\[
P(X > 30) = 1 - 0.1119 = 0.8881
\]

Thus, the probability that the mean time is over 30 seconds is approximately **0.8881**.

### **b. Probability that the mean time is under 35 seconds**

Now, we calculate the z-score for a mean time of 35 seconds:

\[
z = \frac{35 - 32}{1.639} \approx 1.831
\]

For \( z = 1.831 \), the cumulative probability is approximately 0.9666. Since we are looking for the probability that the mean time is **under** 35 seconds, we use this value directly:

\[
P(X < 35) = 0.9666
\]

Thus, the probability that the mean time is under 35 seconds is approximately **0.9666**.

### **c. Probability that the mean time is under 30 seconds or over 35 seconds**

We can find this by calculating:

\[
P(X < 30 \text{ or } X > 35) = P(X < 30) + P(X > 35)
\]

From part (a), we already found \( P(X > 30) = 0.8881 \), so \( P(X < 30) = 1 - 0.8881 = 0.1119 \).

From part (b), \( P(X < 35) = 0.9666 \), so \( P(X > 35) = 1 - 0.9666 = 0.0334 \).

Now we sum the two probabilities:

\[
P(X < 30 \text{ or } X > 35) = 0.1119 + 0.0334 = 0.1453
\]

Thus, the probability that the mean time is under 30 seconds or over 35 seconds is approximately **0.1453**.

### Summary of Results:

- **a.** Probability the mean time is over 30 seconds: **0.8881**
- **b.** Probability the mean time is under 35 seconds: **0.9666**
- **c.** Probability the mean time is under 30 seconds or over 35 seconds: **0.1453**

---

Would you like any additional details or have any questions? Here are some related questions to explore:

1. What is the probability that the mean time is exactly 32 seconds?
2. How does increasing the sample size to 100 travellers affect the probability in part (a)?
3. What would be the probability that the mean time is between 30 and 35 seconds?
4. How is the standard error affected if the sample size is reduced to 25 travellers?
5. Can you explain how the Central Limit Theorem applies to this problem?

**Tip:** The larger your sample size, the smaller the standard error, which means the sample mean tends to be closer to the population mean.

The amount of time travellers at an airport spend with customs officers has a mean of μ = 32 seconds and a standard deviation of σ = 11 seconds. For a random sample of 45 travellers, what is the probability that their mean time spent with customs officers will be: a. Over 30 seconds? b. Under 35 seconds? c. Under 30 seconds or over 35 seconds?

Solve probability problems using the Central Limit Theorem and normal distribution for a sample of travellers at an airport. Calculate the probability that the mean time spent with customs officers is over 30 seconds, under 35 seconds, or under 30 seconds or over 35 seconds. This solution covers key statistical concepts like Z-scores and standard error.

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