Help: normal distribution question

6000 eggs in a large store have masses which are normally distributed. The mean mass of the eggs are 42 grams with a standard deviation of 5.5 grams.

1) 4000 of the 6000 eggs have masses greater than W grams. Estimate the value of W correct to one decimal place??

Thanks in advance - I can't solve it

Oh and The answer in book is 39.6 grams.

I have tried the Z score formula:
z= x- mean all over standard deviation
but can't answer it still??

You should have showed what you tried! The "standard" normal distribution is the \(\displaystyle \frac{x- \mu}{\sigma}\), the variable minus the mean, the divided by the standard deviation.

Here that would be \(\displaystyle z= \frac{W- 42}{5.5}\). Perhaps the difficulty is that you are asked to do this "in reverse"- that is, given the probability: \(\displaystyle P(z< Z)= \frac{4000}{6000}= \frac{2}{3}= 0.666...\). Use a table of the "standard normal distribution" (there's certainly one in your book but a good one is at http://www.mathsisfun.com/data/standard-normal-distribution-table.html) to find z such that the area under the curve is equal to 0.66 (The lower half is, of course, 0.50 so if your table only give z> 0, use .66- .50= .16.). Set that equal to \(\displaystyle \frac{W- 42}{5.5}\) and solve for W.

Thanks for reply-
I understand workin with the usual Z score and if I got a negative answer I could turn it around and use 1 - the positive of it. I don't understand why we change it to {mean - x} now though?

Z= 0.43 for the area 0.6666

0.43 = x-42 / 5.5
Which works out as x = 44.4.
i can't get my head around the formula been changed

Stephencormac said:

Z= 0.43 for the area 0.6666

0.43 = x-42 / 5.5
Which works out as x = 44.4.
i can't get my head around the formula been changed

Click to expand...

To help get your had around it, draw a sketch of a bell curve. Since it is symmetric, half of the area is above and half is below the mean. In your problem, you are looking for the point on the curve such that 2/3 of the area is above and 1/3 is below. Make an eyeball estimate of where the lower tail is about 1/3 of the total area, and shade that fraction. Now when you look it up in the table and find the relevant z-value is 0.43, you should realize that you are looking on the the left-side tail - that is, you need to find 0.43 standard deviations less than the mean.

-0.43 = (x - 42) / 5.5