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

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* 0.1=20.2 ounce`

The company re-calibrates the machine

`P\left ( \bar{x}> 20.2\right )=P\left ( z-(20.2-20)/((1)/(√(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* 0.1587=16\ days`