Question 1165346
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import norm

# Population parameters
mu = 4.2
sigma = 0.6
n = 38

# Sample mean of interest
x_bar = 3.9

# 1. Calculate the standard error of the mean (sigma_x_bar)
sigma_x_bar = sigma / np.sqrt(n)

# 2. Calculate the Z-score
Z = (x_bar - mu) / sigma_x_bar

# 3. Calculate the probability P(M <= 3.9) using the cumulative distribution function (CDF)
probability = norm.cdf(Z)

print(f"Standard Error (sigma_x_bar): {sigma_x_bar}")
print(f"Z-score: {Z}")
print(f"Probability P(M <= 3.9): {probability}")

```

```text?code_stdout&code_event_index=2
Standard Error (sigma_x_bar): 0.09733285267845752
Z-score: -3.082207001484491
Probability P(M <= 3.9): 0.0010273594880658952

```

The problem asks for the probability that the mean replacement time ($\bar{X}$) of 38 randomly selected laptops is $3.9$ years or less, given the population mean ($\mu$) is $4.2$ years and the population standard deviation ($\sigma$) is $0.6$ years.

Since the sample size ($n=38$) is large enough ($n > 30$), we can use the Central Limit Theorem (CLT).

### 1\. Calculate the Standard Error ($\sigma_{\bar{x}}$)

The standard deviation of the sample means is the standard error:
$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.6}{\sqrt{38}}$$
$$\sigma_{\bar{x}} \approx 0.09733$$

### 2\. Calculate the Z-score

The $Z$-score measures how many standard errors the sample mean is from the population mean:
$$Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{3.9 - 4.2}{0.09733}$$
$$Z = \frac{-0.3}{0.09733} \approx -3.0822$$

### 3\. Find the Probability $P(\bar{X} \le 3.9)$

We look up the area to the left of $Z \approx -3.0822$ in the standard normal table.

$$P(Z \le -3.0822) \approx 0.001027$$

Rounding the answer to 4 decimal places:

$$P(\bar{X} \le 3.9 \text{ years}) = \mathbf{0.0010}$$