Question

according to a study, the time between taking a pain reliever and getting relief for a randomly selected patient has unknown distribution with a mean of 49 minutes and a standard deviation of 5.3 minutes. let (x) be the time between taking a pain reliever and getting relief for a randomly selected patient and let (\bar{x}) be the average time between taking a pain reliever and getting relief for a random sample of size 33. 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) unknown ((mu = 49,sigma = 5.3)) and find the probability that the time between taking a pain reliever and getting relief for a randomly selected patient is between 53 and 58 minutes. dne (round the answer to 4 decimal places) 2. use the central limit theorem the sample size is large ((n > 30)) although the distribution of the original population is unknown to describe the probability distribution of (\bar{x}) and state its parameters (mu_{\bar{x}}) and (sigma_{\bar{x}}): (round the answers to 1 decimal place) (\bar{x}sim n(mu_{\bar{x}} = 49,sigma_{\bar{x}} = 9)) and find the probability that the average time between taking a pain reliever and getting relief for a sample of 33 randomly selected patients is more than 50 minutes. (round the answer to 4 decimal places)

Explanation:

Step1: Recall Central Limit Theorem

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 when $n$ is large ($n\geq30$). Here, $\mu = 49$, $\sigma = 5.3$ and $n = 33$.

Step2: Calculate the standard deviation of the sample mean

The formula for the standard deviation of the sample mean (also known as the standard error) is $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Substituting $\sigma = 5.3$ and $n = 33$, we get $\sigma_{\bar{X}}=\frac{5.3}{\sqrt{33}}\approx0.9$.

Step3: Standardize the value

We want to find $P(\bar{X}>50)$. First, we standardize the value of $\bar{X}$ using the formula $z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. Here, $\mu_{\bar{X}}=\mu = 49$ and $\sigma_{\bar{X}}\approx0.9$. So, $z=\frac{50 - 49}{0.9}\approx1.11$.

Step4: Find the probability

We know that $P(\bar{X}>50)=P(Z > 1.11)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq1.11) = 0.8665$, then $P(Z>1.11)=1 - 0.8665 = 0.1335$.

Answer:

$0.1335$