Consider a population with mean u = 50 and standard deviation o = 4. Let X1,X2, ... ,X10 be a random sample and let X the sample mean. Find the expected value and standard deviation of X. Do you know the probability distribution of X? Clearly state any necessary assumption.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![**Title: Exploring the Sample Mean in a Population**
**Introduction:**
Consider a population with a mean (μ) of 50 and a standard deviation (σ) of 4. A random sample, denoted as \( X_1, X_2, ..., X_{10} \), is taken from this population. Here, we explore the sample mean \( \overline{X} \).
**Objective:**
- Determine the expected value and standard deviation of the sample mean \( \overline{X} \).
- Identify the probability distribution of \( \overline{X} \).
**Analysis:**
1. **Expected Value of \( \overline{X} \):**
- The expected value of the sample mean \( \overline{X} \) is equal to the population mean, which is 50.
- \( E(\overline{X}) = \mu = 50 \).
2. **Standard Deviation of \( \overline{X} \):**
- The standard deviation of the sample mean is calculated using the formula:
\[
\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}
\]
- Where \( n \) is the sample size (10 in this case).
- \( \sigma_{\overline{X}} = \frac{4}{\sqrt{10}} \approx 1.26 \).
3. **Probability Distribution of \( \overline{X} \):**
- According to the Central Limit Theorem, the distribution of the sample mean \( \overline{X} \) will approximate a normal distribution if the sample size is sufficiently large (usually \( n \geq 30 \) is considered large, but \( n = 10 \) is often sufficient especially if the original population distribution is normal).
- Hence, \( \overline{X} \) follows a normal distribution with:
- Mean (\( \mu_{\overline{X}} \)) = 50
- Standard deviation (\( \sigma_{\overline{X}} \)) = 1.26
**Conclusion:**
To apply these findings, assume the population distribution is normal or \( n \) is large enough for the Central Limit Theorem to hold. This ensures that \( \overline{X} \) can be treated as normally distributed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ba8d2f4-1578-41b8-b06f-5701357abb5d%2F58117d94-6ecf-479d-93ae-6a0f5b8c3fe3%2Fwyqchcd_processed.png&w=3840&q=75)
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