Question

A manufacturer of cereal has a machine that, when working properly, puts 20 ounces of cereal on average into a box with a standard deviation of 1 ounce. Every morning workers weigh 25 filled boxes. If the average weight is off by more than 1 percent from the desired 20 ounces per box, company policy requires them to recalibrate the machine. In a sample of 100 days where the machine is working properly all day, on how many of the days is it expected the machine will be recalibrated?

Answer 1

Answer:

Step-by-step explanation:

Given

mean $\mu =20\ ounces$

standard deviation $\sigma =1\ ounce$

The no of sample boxes weigh Every morning is 25

Average weight is 1 % more than average

i.e. $20+20\times 0.1=20.2 ounce$

The company re-calibrates the machine

$P\left ( \bar{x}> 20.2\right )=P\left ( z-\frac{20.2-20}{\frac{1}{\sqrt{25}}}\right )$

$=P\left ( z> 1\right )=1-P\left ( z\leq 1\right )$

$=1-0.8413$

$=0.1587$

Therefore the Probability that the average weight of box is more than 20.2 ounce is 0.1587

No of days the machine is expected to re-calibrate is $100\times 0.1587=16\ days$