Normal Distribution question

The weight of eggs measured in grams can be modelled by a X~N(85, 0.36) distribution. Eggs are classified as large , medium or small, where a large egg weighs 90g or more and 25% of egges are classified as small.
(a) Determine the percentage of eggs which are classified as large.
(b) The Max. weight of a small egg.

Reply 1

BCHL85

Use Z = (X - m)/sqrt(v) = (X - 85)/rt(0.36)
So Z ~ N(0, 1)

a) P(X >= 90) = P(Z > 5/0.6) = 1 - P(Z < 8.33)

b) P(X <= x) = 25% where x is the max weight of a small egg.
P(Z < (x - 85)/0.6) = 0.25
Use the table to find the value for z = (x-85)/0.6

Reply 2

xlaser31

How is part (a) finished off to give a value of 20.2%

Reply 3

Morbo

I don't think there should be 20.2% of eggs classified as large given that for an egg to be large, it has to deviate from the mean by 5g, and the standard deviation is only 0.6g. At 5g, you are at 8.3 standard deviations. The Z-distribution goes to practically zero after about 3 standard deviations!

Either you have provided the incorrect distribution/definition of large egg, or the answer you have is wrong. I'd expect 0% of eggs to be large given your data.

Reply 4

Libertine

^^What he said. 3.6 looks like it should give the right answer...

Reply 5

Morbo

Lol. Yeah, actually that's a more sensible suggestion, Libertine. Chickens are pretty damn accurate if they eggs of 85.00g +/- 0.36g!

Reply 6

Expression

xlaser31

How is part (a) finished off to give a value of 20.2%

Okey I assume you've got that as a published answer, so we'll work on that being the correct information.

If 20.2 % of the eggs are large, then using inverse normal tables, z=0.8345.

So we'd have 0.8345 = 5/sigma

Which gives sigma = 5.9916.... and sigma² = 35.899 (approx 36).

In which case we've got an X~N(85, 36) - the problem with that is that if we do the calculation with that information from the offset, then we produce z=0.833... which gives us a final answer of 20.25 % large, close but not exactly what you've got.

b) Now, if X~N(85, 36) is indeed your distribution then you need to recalculate the value for the small eggs.

Using Inverse Normal Tables, and symmetry of Normal Distribution we find that for 25 % to be small, the z-value at this critical boundary is -0.6745.

Now using the standardizing formula trying to find x:

-0.6745 = (x-85)/6

Rearranging for x; we find x = 80.953g is the largest possible egg to be classified as small.