The number of unbroken charcoal briquets in a 20-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

Given,

$\mu =450,\sigma =20$

USe the Formula :

$Z=\frac{X-\mu }{\sigma }$

Consider,

The book's Z table can be used to compute the 5th percentile; look for the associated Z value within the table.

$$\begin{gathered} Z=\frac{X-\mu }{\sigma } \\ X=\sigma *Z+\mu \\ X=-1.645*20+450=417.1 \end{gathered}$$

Knowing that $Z=-1.645$

$X=418$

Therefore, the correct option is (c)

a. The minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free is not $404$

b. The minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free is not $411$

d. The minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free is not $425$

e. The minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free is not $448.$