Question 1198355: For problem 1 involving the Central Limit Theorem, show complete manual solutions.
1. Suppose that the daily fluctuation in the price of a company stock is a random variable X that is uniformly distributed on the interval -1 to 1. Suppose that a broker observes 216 trading days during the past year. Find the approximate probability by the Central Limit Theorem that the
absolute value of the average of the 216 stock price fluctuations is within 0.02 units away from the mean of the sampling distribution of the sample mean. 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 ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! To solve this problem using the Central Limit Theorem (CLT), we calculate the probability that the sample mean of \( n = 216 \) stock price fluctuations lies within 0.02 units of the mean. Here's the step-by-step process:
---
### **Step 1: Identify the Distribution and Parameters**
1. The random variable \( X \) (daily fluctuation in stock price) is uniformly distributed over \([-1, 1]\).
- Mean (\( \mu \)):
\[
\mu = \frac{a + b}{2} = \frac{-1 + 1}{2} = 0
\]
- Variance (\( \sigma^2 \)):
\[
\sigma^2 = \frac{(b - a)^2}{12} = \frac{(1 - (-1))^2}{12} = \frac{4}{12} = \frac{1}{3}
\]
2. Sampling distribution of the sample mean \( \bar{X} \):
- Mean of \( \bar{X} \): \( \mu_{\bar{X}} = \mu = 0 \)
- Standard deviation of \( \bar{X} \) (Standard Error):
\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{\frac{1}{3}}}{\sqrt{216}} = \frac{\sqrt{1/3}}{14.7} \approx 0.0483
\]
---
### **Step 2: Define the Problem**
We want to find the probability that \( |\bar{X} - \mu| \leq 0.02 \), which can be expressed as:
\[
P(-0.02 \leq \bar{X} - \mu \leq 0.02)
\]
Since \( \bar{X} \) is approximately normal (by the CLT), this translates to:
\[
P\left( -\frac{0.02}{\sigma_{\bar{X}}} \leq Z \leq \frac{0.02}{\sigma_{\bar{X}}} \right)
\]
where \( Z \) is the standard normal variable.
---
### **Step 3: Standardize the Interval**
Using \( \sigma_{\bar{X}} \approx 0.0483 \), compute the \( Z \)-scores:
\[
Z_1 = \frac{-0.02}{0.0483} \approx -0.414
\]
\[
Z_2 = \frac{0.02}{0.0483} \approx 0.414
\]
---
### **Step 4: Find the Probability**
Using the standard normal table or a calculator, find the probabilities corresponding to \( Z_1 = -0.414 \) and \( Z_2 = 0.414 \):
\[
P(Z \leq 0.414) \approx 0.6603
\]
\[
P(Z \leq -0.414) \approx 1 - P(Z \leq 0.414) = 0.3397
\]
The probability that \( Z \) is between \( -0.414 \) and \( 0.414 \) is:
\[
P(-0.414 \leq Z \leq 0.414) = P(Z \leq 0.414) - P(Z \leq -0.414)
\]
\[
P(-0.414 \leq Z \leq 0.414) = 0.6603 - 0.3397 = 0.3206
\]
---
### **Step 5: Final Answer**
The approximate probability that the sample mean is within 0.02 units of the mean is:
\[
P(|\bar{X} - \mu| \leq 0.02) \approx 0.3206 \, (32.06\%).
\]
---
### **Flowchart of Clouds**
Here’s the structure of the flowchart for the Central Limit Theorem (CLT):
1. **Cloud 1**: Population distribution \( X \sim U(-1, 1) \)
- Mean \( \mu = 0 \)
- Variance \( \sigma^2 = \frac{1}{3} \)
2. **Cloud 2**: Sampling distribution of the sample mean \( \bar{X} \)
- Sample size \( n = 216 \)
- Mean \( \mu_{\bar{X}} = \mu = 0 \)
- Standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \approx 0.0483 \)
3. **Cloud 3**: Approximation of the sampling distribution \( \bar{X} \) by the normal distribution via CLT:
- \( \bar{X} \sim N(0, 0.0483^2) \)
4. **Cloud 4**: Standard normal distribution \( Z \sim N(0, 1) \), used for probability calculations.
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