SOLUTION: 1. A business owner has designed a pay scale for his employees that follow this distribution, f(x) = (20-x)/50, 10 <= x <= 20. The mean hourly rate at this business is $13.33. If a

Question 1162247: 1. A business owner has designed a pay scale for his employees that follow this distribution, f(x) = (20-x)/50, 10 <= x <= 20. The mean hourly rate at this business is $13.33. If an auditor randomly selects 20 employees from the company, how likely is she to observe an AVERAGE hourly rate above $15.

2. The variable x comes from the mgf, M(t) = 1/(1-nt)^alpha, where n = 2, alpha = 11.5. Find the probability that x is between 14.848 and 32.007.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=1
import numpy as np
import scipy.stats as stats
# Problem 1: Sampling distribution of the mean
# f(x) = (20-x)/50, 10 <= x <= 20
# E[X] = 13.33 (given)
# Var(X) = E[X^2] - (E[X])^2
# E[X^2] = Integral from 10 to 20 of x^2 * (20-x)/50 dx
# E[X^2] = 1/50 * [20*x^3/3 - x^4/4] from 10 to 20
e_x2 = (1/50) * ((20 * 20**3 / 3 - 20**4 / 4) - (20 * 10**3 / 3 - 10**4 / 4))
mu = 13.333333333
var_x = e_x2 - mu**2
std_x = np.sqrt(var_x)
# CLT parameters
n_samp = 20
mu_xbar = mu
std_xbar = std_x / np.sqrt(n_samp)
# P(X_bar > 15)
prob_1 = 1 - stats.norm.cdf(15, loc=mu_xbar, scale=std_xbar)
# Problem 2: MGF M(t) = (1 - nt)^-alpha
# This is the MGF of a Gamma distribution with parameters:
# shape (k) = alpha, scale (theta) = n
# Note: Some definitions of Gamma MGF use (1 - beta*t)^-alpha
alpha_val = 11.5
n_val = 2.0
# P(14.848 < X < 32.007)
prob_2 = stats.gamma.cdf(32.007, a=alpha_val, scale=n_val) - stats.gamma.cdf(14.848, a=alpha_val, scale=n_val)
print(f"{mu=}, {var_x=}, {std_x=}")
print(f"{std_xbar=}")
print(f"{prob_1=}")
print(f"{prob_2=}")
```
```text?code_stdout&code_event_index=1
mu=13.333333333, var_x=5.555555564444461, std_x=2.35702260584078
std_xbar=0.5270462771163676
prob_1=0.0007827011341015044
prob_2=0.800000547869999
```
### 1. Sampling Distribution of the Pay Scale
To find the probability that the average hourly rate of 20 employees is above $15, we use the **Central Limit Theorem (CLT)**.
**Step 1: Find the Variance of the Population**
We are given the PDF $f(x) = \frac{20-x}{50}$ for $10 \leq x \leq 20$ and the mean $\mu = 13.33$. We need the variance $\sigma^2$:
$$\sigma^2 = E[X^2] - (E[X])^2$$
$$E[X^2] = \int_{10}^{20} x^2 \left(\frac{20-x}{50}\right) dx = \frac{1}{50} \left[ \frac{20x^3}{3} - \frac{x^4}{4} \right]_{10}^{20} \approx 183.33$$
$$\sigma^2 = 183.33 - (13.33)^2 \approx 5.56 \implies \sigma \approx 2.36$$
**Step 2: Parameters for the Sample Mean ($\bar{x}$)**
* **Mean of $\bar{x}$:** $\mu_{\bar{x}} = 13.33$
* **Standard Error:** $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.36}{\sqrt{20}} \approx 0.527$
**Step 3: Calculate the Z-score and Probability**
$$z = \frac{15 - 13.33}{0.527} \approx 3.17$$
Using the standard normal table, $P(Z > 3.17) \approx 0.00078$.
**Conclusion:** It is highly unlikely (**0.078%**) that the average rate would be above $15.
---
### 2. Probability from the Moment Generating Function (MGF)
The MGF $M(t) = \frac{1}{(1-nt)^\alpha}$ is the characteristic form of a **Gamma Distribution** where:
* **Shape ($\alpha$):** 11.5
* **Scale ($\theta$ or $n$):** 2
*(Note: This is also equivalent to a Chi-Squared distribution if $\alpha$ were an integer/half-integer and $n=2$, but here we treat it as a Gamma distribution.)*
**The Calculation:**
We need $P(14.848 < X < 32.007)$ for $X \sim \text{Gamma}(\text{shape}=11.5, \text{scale}=2)$.
Using the Gamma Cumulative Distribution Function (CDF):
* $P(X \leq 32.007) \approx 0.9000$
* $P(X \leq 14.848) \approx 0.1000$
$$P(14.848 < X < 32.007) = 0.9000 - 0.1000 = 0.8000$$
**Conclusion:** The probability is **0.8000** (or **80%**).
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