Question

suppose x has a distribution with μ = 22 and σ = 21. (a) if a random sample of size n = 36 is drawn, find the following. find μ_x̄. μ_x̄ = find σ_x̄. σ_x̄ = find p(22 ≤ x̄ ≤ 24). (round your answer to four decimal places.) p(22 ≤ x̄ ≤ 24) = (b) if a random sample of size n = 64 is drawn, find the following. find μ_x̄. μ_x̄ = find σ_x̄. σ_x̄ = find p(22 ≤ x̄ ≤ 24). (round your answer to four decimal places.) p(22 ≤ x̄ ≤ 24) = (c) why should you expect the probability of part (b) to be higher than that of part (a)? (hint: consider the standard deviations in parts (a) and (b).) the standard deviation of part (b) is select part (a) because of the select sample size. therefore, the distribution about μ_x̄ is select

Explanation:

Step1: Calculate $\mu_{\bar{x}}$

According to the Central Limit Theorem, for a sample of size $n$ from a population with mean $\mu$, the mean of the sample mean $\bar{x}$ is $\mu_{\bar{x}}=\mu$. Given $\mu = 22$, so $\mu_{\bar{x}}=22$ for both (a) and (b).

Step2: Calculate $\sigma_{\bar{x}}$ for part (a)

The formula for the standard deviation of the sample mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 21$ and $n = 36$, then $\sigma_{\bar{x}}=\frac{21}{\sqrt{36}}=\frac{21}{6} = 3.5$.

Step3: Calculate $P(22\leq\bar{x}\leq24)$ for part (a)

First, standardize the values. $z_1=\frac{22 - 22}{3.5}=0$ and $z_2=\frac{24 - 22}{3.5}=\frac{2}{3.5}\approx0.57$. Then $P(22\leq\bar{x}\leq24)=P(0\leq Z\leq0.57)$. Using the standard normal table, $P(0\leq Z\leq0.57)=\varPhi(0.57)-\varPhi(0)$. Since $\varPhi(0) = 0.5$ and $\varPhi(0.57)=0.7157$, so $P(0\leq Z\leq0.57)=0.7157 - 0.5=0.2157$.

Step4: Calculate $\sigma_{\bar{x}}$ for part (b)

Using the formula $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$ with $\sigma = 21$ and $n = 64$, we get $\sigma_{\bar{x}}=\frac{21}{\sqrt{64}}=\frac{21}{8}=2.625$.

Step5: Calculate $P(22\leq\bar{x}\leq24)$ for part (b)

Standardize the values. $z_1=\frac{22 - 22}{2.625}=0$ and $z_2=\frac{24 - 22}{2.625}=\frac{2}{2.625}\approx0.76$. Then $P(22\leq\bar{x}\leq24)=P(0\leq Z\leq0.76)$. Using the standard normal table, $\varPhi(0.76)=0.7764$ and $\varPhi(0) = 0.5$. So $P(0\leq Z\leq0.76)=0.7764 - 0.5=0.2764$.

Answer:

(a) $\mu_{\bar{x}} = 22$, $\sigma_{\bar{x}}=3.5$, $P(22\leq\bar{x}\leq24)=0.2157$
(b) $\mu_{\bar{x}} = 22$, $\sigma_{\bar{x}}=2.625$, $P(22\leq\bar{x}\leq24)=0.2764$
(c) The standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about $\mu_{\bar{x}}$ is narrower.