Question

A machine which fills 12 oz cans of soda actually filsl the can a random amount which is normally distrubuted with a mean of 12 oz and a standard deviation of 0.8 oz. What is the cut off point so that 93.0% of the cans will have that much soda or more?

Answer 1

93.0% of the cans will have 10.82 oz of soda or more.

Explanation:

We are given the following information in the question:

Mean, μ = 12 oz

Standard Deviation, σ = 0.8 oz

We are given that the distribution of soda fills is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to find the value of x such that the probability is 0.93.

P(X > x)

`P( X > x) = P( z > \displaystyle(x - 12)/(0.8))=0.93`

`= 1 -P( z \leq \displaystyle(x - 12)/(0.8))=0.93`

`=P( z \leq \displaystyle(x - 12)/(0.8))=0.07`

Calculation the value from standard normal z table, we have,

`P(z<-1.476) = 0.07`

`\displaystyle(x - 12)/(0.8) = -1.476\\x = 10.8192 \approx 10.82`

Hence, 93.0% of the cans will have 10.82 oz of soda or more.