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kari74 [83]
3 years ago
13

A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.1

6 ounces2. If 85% of the containers hold greater than 12 ounces, find the machine's mean filling weight (in ounces).
Mathematics
1 answer:
Ghella [55]3 years ago
8 0

Answer:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sqrt{0.16}= 0.4)  

For this part we want to find a value a, such that we satisfy this condition:

P(X>12)=0.85   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.85 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.85

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

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