The weight of a product is normally distributed with a standard deviation of .8 ounces. What should the average weight be if the production manager wants no more than 5% of the products to weigh more than 5.3 ounces?

Question

The weight of a product is normally distributed with a standard deviation of .8 ounces. What should the average weight be if the production manager wants no more than 5% of the products to weigh more than 5.3 ounces?

Answer 1

Hi Bob,

Thanks for the direction. I got so far with it on my own and became stuck. I did as you said but still have one question. Would the answer to the question be 1.31 ounces or 6.61 ounces? I feel the answer is 1.31. Thanks

Answer 2

To determine the average weight that fulfills the production manager's requirement, we need to express the weight distribution in terms of the standard normal distribution.

Step 1: Find the Z-score

The Z-score represents how many standard deviations a particular weight is from the mean of the weight distribution. It is calculated using the formula:

Z = (X - μ) / σ

where Z is the Z-score, X is the observed weight, μ is the mean of the weight distribution, and σ is the standard deviation of the weight distribution.

In this case, we have:
X = 5.3 ounces
σ = 0.8 ounces

Step 2: Find the corresponding percentile

Since we want no more than 5% of the products to weigh more than 5.3 ounces, we are looking for the value on the standard normal distribution that corresponds to a percentile of 95% (100% - 5%).

We can find this value using a standard normal distribution table or a statistical calculator. The value corresponding to a 95% percentile is approximately 1.645.

Step 3: Solve for the mean

Now that we have the Z-score and corresponding percentile, we can rearrange the Z-score formula to solve for the mean (μ):

Z = (X - μ) / σ

rearranges to:

μ = X - (Z * σ)

Substituting the known values, we have:

μ = 5.3 - (1.645 * 0.8)

Calculating this expression:

μ ≈ 5.3 - 1.316

Therefore, the average weight should be approximately 3.984 ounces (rounded to three decimal places) to ensure that no more than 5% of the products weigh more than 5.3 ounces.

Answer 3

http://davidmlane.com/hyperstat/z_table.html

second applet.
Set mean to zero, std deviation to .8 oz
check below
enter shaded area .05

Notice the answer comes up 1.31 ounces. So the mean must be 5.3+1.31=6.61

Put that in for the mean, and verify.

You should be able to do with with your calculator also, grab your book and read. Finally, you may be learning to do this with tables...for that, hump, grunt, and strain through it.