SOLUTION: For problem 2 involving the Central Limit Theorem, show complete manual solutions. 2. a. (This part does not require a computer solution) Find the value of the constant c

Question 1198356: For problem 2 involving the Central Limit Theorem, show complete manual solutions.
2. a. (This part does not require a computer solution) Find the value of the constant c so that the following function is the probability density function of the random variable X that is normally distributed. Identify the mean and variance of X.
𝑓(𝑥)=𝑐𝑒^−4𝑥2−16𝑥−16,−∞<𝑥 <∞
(This part requires a computer solution) In addition, find the probability 𝑃(0.8<𝑋 <2)

b. (This part requires a computer solution) If many random samples of size 25 are taken from the distribution in part (2a), what would the mean and the variance of the sampling distribution of the sample mean 𝑋̅ be expected to equal? What distribution would the sample means follow? Determine the 95th percentile of the sampling distribution of 𝑋̅ and give a valid interpretation of what the value means. Since this is a CLT problem, draw the “flowchart of clouds” as demonstrated in the lectures with all the elements of the CLT and show complete solutions.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
### Problem 2a: Finding the Constant \( c \), Mean, and Variance of \( X \)
The probability density function \( f(x) \) is given by:
\[
f(x) = c e^{-4x^2 - 16x - 16}, \quad -\infty < x < \infty
\]
#### **Step 1: Rewrite the exponent in quadratic form**
The exponent \( -4x^2 - 16x - 16 \) can be rewritten by completing the square:
\[
-4x^2 - 16x - 16 = -4(x^2 + 4x + 4) = -4(x + 2)^2
\]
Thus, the probability density function becomes:
\[
f(x) = c e^{-4(x + 2)^2}
\]
#### **Step 2: Normalize the PDF**
The integral of \( f(x) \) over all \( x \) must equal 1:
\[
\int_{-\infty}^\infty f(x) \, dx = 1
\]
\[
\int_{-\infty}^\infty c e^{-4(x + 2)^2} \, dx = 1
\]
Use the Gaussian integral property:
\[
\int_{-\infty}^\infty e^{-a(x+b)^2} \, dx = \sqrt{\frac{\pi}{a}}
\]
Here, \( a = 4 \), so:
\[
\int_{-\infty}^\infty e^{-4(x + 2)^2} \, dx = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}
\]
Substitute into the normalization equation:
\[
c \cdot \frac{\sqrt{\pi}}{2} = 1 \quad \implies \quad c = \frac{2}{\sqrt{\pi}}
\]
#### **Step 3: Identify the mean and variance**
The PDF \( f(x) = \frac{2}{\sqrt{\pi}} e^{-4(x + 2)^2} \) is a normal distribution with:
- Mean (\( \mu \)): The center of the quadratic term \( -4(x + 2)^2 \) is at \( x = -2 \), so \( \mu = -2 \).
- Variance (\( \sigma^2 \)): The coefficient of \( (x + 2)^2 \) is \( -4 = -\frac{1}{2\sigma^2} \), so \( \sigma^2 = \frac{1}{8} \).
#### **Final Answer for Part 2a**:
- Constant \( c = \frac{2}{\sqrt{\pi}} \)
- Mean (\( \mu \)) = \( -2 \)
- Variance (\( \sigma^2 \)) = \( \frac{1}{8} \)
---
### Probability \( P(0.8 < X < 2) \)
We compute:
\[
P(0.8 < X < 2) = \int_{0.8}^2 f(x) \, dx
\]
Substituting \( f(x) = \frac{2}{\sqrt{\pi}} e^{-4(x + 2)^2} \), this requires numerical integration. Let me calculate this using Python.
The probability \( P(0.8 < X < 2) \) is approximately \( 0 \). This result suggests that the values \( 0.8 \) and \( 2 \) are far from the mean (\( -2 \)) relative to the small variance (\( \frac{1}{8} \)), making the probability negligible.
---
### Part 2b: Sampling Distribution of \( \bar{X} \)
#### **Mean and Variance of \( \bar{X} \):**
If random samples of size \( n = 25 \) are taken:
- The mean of the sampling distribution (\( \mu_{\bar{X}} \)) equals the population mean:
\[
\mu_{\bar{X}} = \mu = -2
\]
- The variance of the sampling distribution (\( \sigma^2_{\bar{X}} \)) is:
\[
\sigma^2_{\bar{X}} = \frac{\sigma^2}{n} = \frac{\frac{1}{8}}{25} = \frac{1}{200} = 0.005
\]
Hence, the standard deviation is:
\[
\sigma_{\bar{X}} = \sqrt{0.005} \approx 0.0707
\]
#### **Distribution of \( \bar{X} \):**
By the Central Limit Theorem, \( \bar{X} \) follows a normal distribution:
\[
\bar{X} \sim N(-2, 0.0707^2)
\]
---
#### **95th Percentile of the Sampling Distribution:**
The 95th percentile corresponds to \( z = 1.645 \) for the upper tail of the standard normal distribution:
\[
\text{95th Percentile} = \mu_{\bar{X}} + z \cdot \sigma_{\bar{X}} = -2 + 1.645 \cdot 0.0707
\]
Let me calculate this.
The 95th percentile of the sampling distribution of \( \bar{X} \) is approximately \( -1.8837 \).
#### **Interpretation:**
If many samples of size 25 are taken, 95% of the sample means will be less than \( -1.8837 \).
---
### Flowchart of Clouds
- **Population:** \( N(-2, 0.125) \), where \( \mu = -2 \) and \( \sigma^2 = \frac{1}{8} \).
- **Sampling Distribution:** \( \bar{X} \sim N(-2, 0.005) \), with \( \sigma_{\bar{X}} = 0.0707 \).
This demonstrates the Central Limit Theorem by showing how the sampling distribution narrows (smaller variance) as sample size increases.
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