The number of unbroken charcoal briquets in a twenty-pound bag filled at the factory follows a Normal distribution with a mean of$450$briquets and a standard deviation of $20$briquets. The company expects that a certain number of the bags will be underfilled, so the company will replace for free the 5% of bags that have too few briquets. What is the minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free?

(a) $404$

(b) $411$

(c) $418$

(d) $425$

(e)$448$

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.

$X~Normal(\mu =450,\sigma =20)$

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

$Z~Normal(\mu =0,\sigma =1)$

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" $$\begin{gathered} Z=\frac{X-\mu }{\sigma } \\ X=\sigma *Z+\mu \end{gathered}$$

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).

$X=418.$