91Ó°ÊÓ

Given that the weight and the standard deviation of the sample weights of candies in the table.

$Meansampleweight=AVERAGE(A2â‹\dots A14,B2â‹\dots B26,C2â‹\dots C9,D2â‹\dots D9,E2â‹\dots E27,F2â‹\dots F20)$

$S\tan darddeviationofsampleweights=\mathrm{STDEV}â‹\dots \mathrm{S}(\mathrm{A}2â‹\dots \mathrm{A}14,\mathrm{B}2â‹\dots \mathrm{B}26,\mathrm{C}2:\mathrm{C}9,\mathrm{D}2â‹\dots \mathrm{D}9,\mathrm{E}2â‹\dots \mathrm{E}27,\mathrm{F}2â‹\dots \mathrm{F}20)$

And the result is

$Meansampleweight=0.856191919$

$S\tan darddeviationofsampleweights=0.051971475$

M&M candies large candy bags have a claimed net weight of $396.6g$. The standard deviation for the weight of the individual candies is$0.017g$$0.017g$

To estimate the sample weights and the standard deviation, utilize the formula

Sum of the sample weight$=SUM(A2:A14,B2:B26,C2:C9,D2:D9,E2:E27,F2:F20)$

Standard deviation of sum of the weight $=H4×100$

Therefore, the sum of the sample weight is $84.763$

The standard deviation of the sum of the weight is$5.197147506$

Given that $465$M&Ms are randomly selected, the probability that the sum of their weights to at least $396.9$.

To ascertain the probability that the weighted total is no less than $396.9$ pounds, utilize the $Ti-83$calculator, for this, click on $2^{\text{nd}}$, then DISTR, and afterward look down to the ordinary CDF choice and enter the given subtleties. After this, click on ENTER button of the number cruncher to have the ideal outcome.

$$\begin{gathered} normalcdf(396.9,E99,(465\left)\right(0.85619),(\sqrt{}\left(465\right)\left)\right(0.0519\left)\right) \\ -.1361983597 \end{gathered}$$

Hence the probability is

$1-0.1361=0.8639$

Given that Mars Company’s M&M labeling.

The probability of weight sum at least $396.9$ pounds is approximately $0.8639$. So marsh is correctly labeling their $MM$ packages.