Question

this year, the act score of a randomly selected student has unknown distribution with a mean of 23.7 points and a standard deviation of 6 points. let (x) be the act score of a randomly selected student and let (overline{x}) be the average act score of a random sample of size 46. note: if a probability cannot be determined, enter dne (does not exist). 1. describe the probability distribution of (x) and state its parameters (mu) and (sigma): (xsim) select an answer ((mu=), (sigma =)) and find the probability that the act score of a randomly selected student is less than 16 points. (round the answer to 4 decimal places) 2. use the central limit theorem select an answer to describe the probability distribution of (overline{x}) and state its parameters (mu_{overline{x}}) and (sigma_{overline{x}}): (round the answers to 1 decimal place) (overline{x}sim) select an answer ((mu_{overline{x}}=), (sigma_{overline{x}}=)) and find the probability that the average act score of a sample of 46 randomly selected students is more than 23 points. (round the answer to 4 decimal places) question help: written example

Explanation:

Step1: Identify parameters for $X$

For a single - randomly selected student's ACT score $X$, the mean $\mu = 23.7$ and the standard deviation $\sigma=6$. Since the distribution of $X$ is unknown, we cannot use the normal - distribution formula directly to find $P(X < 16)$ without further assumptions. But if we assume a normal distribution (for the sake of using the standard normal $z$ - score formula), the $z$ - score is calculated as $z=\frac{x-\mu}{\sigma}$.
$z=\frac{16 - 23.7}{6}=\frac{-7.7}{6}\approx - 1.2833$
$P(X < 16)=P(Z<-1.2833)$
Using the standard normal table, $P(Z < - 1.2833)\approx0.1003$

Step2: Apply Central Limit Theorem for $\bar{X}$

The Central Limit Theorem states that for a sample of size $n$ from any population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}}=\mu$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$.
Given $\mu = 23.7$, $\sigma = 6$, and $n = 46$.
$\mu_{\bar{X}}=\mu=23.7$
$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{6}{\sqrt{46}}\approx\frac{6}{6.7823}\approx0.9$
To find $P(\bar{X}>23)$, we first calculate the $z$ - score:
$z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{23 - 23.7}{0.9}=\frac{-0.7}{0.9}\approx - 0.78$
$P(\bar{X}>23)=1 - P(\bar{X}\leq23)=1 - P(Z\leq - 0.78)$
Using the standard - normal table, $P(Z\leq - 0.78) = 0.2177$, so $P(\bar{X}>23)=1 - 0.2177 = 0.7823$

1. $X\sim$ (unknown distribution) $(\mu = 23.7,\sigma = 6)$; $P(X < 16)\approx0.1003$
2. $\bar{X}\sim N(\mu_{\bar{X}} = 23.7,\sigma_{\bar{X}}\approx0.9)$; $P(\bar{X}>23)\approx0.7823$

Answer:

1. $X\sim$ (unknown distribution) $(\mu = 23.7,\sigma = 6)$; $0.1003$
2. $\bar{X}\sim N(\mu_{\bar{X}} = 23.7,\sigma_{\bar{X}}\approx0.9)$; $0.7823$