Question

6 O-10 points JKES1 7 E 034 The local bakery bakes more than a thousand 1 pound loaves of bread daily, and the weights of these loves varies. The mean weight is 1。b. and 4 oz.. or 97S grams. Assume the standard deviation of the weights is 28 grams and a sample of 45 loaves is to be randomly selected (a) This sample of 45 has a mean value of , which belongs to a sampling distribution. Find the shape of this sampling distribution O skewed right O approximately normal skewed left chi-square ) Find the mean of this sampling distribution. (Glve your answer correct to nearest whele (c) Find the standard error of this sampling distribution (Give your answer oorrect to two decimal places (d) What is the probability that this sample mean will be between 96s and 985? (Give your answer correct to four decimal places.) grams decimal places.) (e) What is the probability that the sample mean wil have a value less than 9727 (Give your answer correct to four that the sample mean will be within 4 grams of the mean? (Give your answer correct to four decimal places.)

Answer 1

$\mathbf{Solution}$

$\mathbf{given:}\;n=45,\;\mu=975,\;\sigma=28$

$\mathbf{formula:}\;z=\frac{\bar x -\mu}{\frac{\sigma}{\sqrt{n}}}$

$\mathbf{{\color{Blue} (a)}}\;{\color{Red} \mathbf{approximately\;normal}}$

$\mathbf{{\color{Blue} (b)}}\;mean \;of\;the \;sampling\;distribution\;(\mu_{\bar x})=\mathbf{{\color{Red} 975}}$

$\\\mathbf{{\color{Blue} (c)}}\;standard\;deviation \;of\;the \;sampling\;distribution\;(\sigma_{\bar x})$

$=\frac{\sigma }{\sqrt{n}}=\frac{28}{\sqrt{45}}=\mathbf{{\color{Red} 4.17}}$

$\\\mathbf{{\color{Blue} (d)}}\;P(965<\bar x<985)\Rightarrow P\left ( \frac{965-975}{\frac{28}{\sqrt{45}}} <\frac{\bar x-\mu}{\frac{\sigma }{\sqrt{n}}}<\frac{985-975}{\frac{28}{\sqrt{45}}} \right )$

$=P(-2.40<z<2.40)$

$=P(z<2.40)-P(z<-2.40)$

$from\;z-table$

$=0.9918-0.0082$

$=\mathbf{{\color{Red} 0.9836}}$

$\\\mathbf{{\color{Blue} (e)}}\;P(\bar x<972)\Rightarrow P\left ( \frac{\bar x-\mu}{\frac{\sigma }{\sqrt{n}}}<\frac{972-975}{\frac{28}{\sqrt{45}}} \right )$

$=P(z<-0.72)$

$from\;z-table$

$=\mathbf{{\color{Red} 0.2358}}$

$\\\mathbf{{\color{Blue} (f)}}\;within\;4\;grams\;of\;the\;mean\;\Rightarrow P(971<\bar x<979)$

$=P\left ( \frac{971-975}{\frac{28}{\sqrt{45}}} <\frac{\bar x-\mu}{\frac{\sigma }{\sqrt{n}}}<\frac{979-975}{\frac{28}{\sqrt{45}}} \right )$

$=P(-0.96<z<0.96)$

$=P(z<0.96)-P(z<-0.96)$

$from\;z-table$

$=0.8315-0.1685$

$=\mathbf{{\color{Red} 0.6630}}$