91Ó°ÊÓ

The number of unbroken charcoal briquets follows a Normal distribution with a mean $450$and standard deviation $20$briquets.

The company will replace $5\%$of bags

Find the minimum number of bags.

The company intends to replace $5\%$of the smallest bags manufactured. The twenty-pound bag is assumed to have a normal distribution with a mean of $450$and a standard deviation of $20$. The purpose is to find the distribution's $5$th percentile.

$X~Normal(\mathrm{μ}=450,\mathrm{σ}=20)$

Because we don't have a normal distribution table with values for a mean of $450$and a standard deviation of $20$, the first step is to obtain the value for a standardised normal distribution: Z

$Z~Normal(\mathrm{μ}=0,\mathrm{σ}=1)$

Use the Z table

$P(Z<-1.645)=0.05;Z=-1.645$

Use the relationship between X and Z to convert our Z value to X

localid="1650258761977" $$\begin{gathered} Z=\frac{X-\mathrm{μ}}{\mathrm{σ}} \\ X=\mathrm{σ}*Z+\mathrm{μ} \end{gathered}$$

From the above values

$X=-1.645*20+450=417.1$

Because you can't have a part number of bags underfilled, round up to $418$(if you round down to $417$, they won't meet the $5$percentile requirement).

$X=418.$