SOLUTION: Let us assume that from a population with mean μ=100 and standard deviation σ=15 a sample random variable of n=900 is selected. What is the probability P(X ̅<101.1)?

Question 1210220: Let us assume that from a population with mean μ=100 and standard deviation σ=15 a sample random variable of n=900 is selected.
What is the probability P(X ̅<101.1)?
What is the probability P(X ̅>101.5)?
What is the probability P(99.3
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let $\bar{X}$ be the sample mean of a random sample of size $n=900$ drawn from a population with mean $\mu=100$ and standard deviation $\sigma=15$.
According to the Central Limit Theorem, the sampling distribution of the sample mean $\bar{X}$ will be approximately normal with mean $\mu_{\bar{X}} = \mu$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.
In this case:
$\mu_{\bar{X}} = 100$
$\sigma_{\bar{X}} = \frac{15}{\sqrt{900}} = \frac{15}{30} = 0.5$
Now we can answer each part of the question by converting the sample mean to a z-score:
$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}$
**1. What is the probability P($\bar{X} < 101.1$)?**
First, calculate the z-score for $\bar{X} = 101.1$:
$z = \frac{101.1 - 100}{0.5} = \frac{1.1}{0.5} = 2.2$
Now, we need to find the probability $P(Z < 2.2)$, where $Z$ is a standard normal random variable. Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z < 2.2) \approx 0.9861$
So, $P(\bar{X} < 101.1) \approx 0.9861$.
**2. What is the probability P($\bar{X} > 101.5$)?**
First, calculate the z-score for $\bar{X} = 101.5$:
$z = \frac{101.5 - 100}{0.5} = \frac{1.5}{0.5} = 3$
Now, we need to find the probability $P(Z > 3)$. This is equal to $1 - P(Z \le 3)$. Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z \le 3) \approx 0.9987$
So, $P(Z > 3) = 1 - 0.9987 = 0.0013$
Thus, $P(\bar{X} > 101.5) \approx 0.0013$.
**3. What is the probability P($99.3 < \bar{X} < 100.8$)?**
First, calculate the z-scores for $\bar{X} = 99.3$ and $\bar{X} = 100.8$:
For $\bar{X} = 99.3$:
$z_1 = \frac{99.3 - 100}{0.5} = \frac{-0.7}{0.5} = -1.4$
For $\bar{X} = 100.8$:
$z_2 = \frac{100.8 - 100}{0.5} = \frac{0.8}{0.5} = 1.6$
Now, we need to find the probability $P(-1.4 < Z < 1.6)$, which is equal to $P(Z < 1.6) - P(Z \le -1.4)$. Looking up the values in a standard normal distribution table or using a calculator, we find:
$P(Z < 1.6) \approx 0.9452$
$P(Z \le -1.4) \approx 0.0808$
So, $P(-1.4 < Z < 1.6) = 0.9452 - 0.0808 = 0.8644$
Thus, $P(99.3 < \bar{X} < 100.8) \approx 0.8644$.
Final Answer: The final answer is $\boxed{P(\bar{X} < 101.1) \approx 0.9861, P(\bar{X} > 101.5) \approx 0.0013, P(99.3 < \bar{X} < 100.8) \approx 0.8644}$
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