# Normal Distribution/Probability problem

I have the following problem:

The weights of large eggs are normally distributed with mean 65 grams and standard deviation 4 grams. The weights of standard eggs are normally distributed with mean 50 grams and standard deviation 3 grams.

1. One large egg and one standard egg are chosen at random. Find the probability that the weight of the standard egg is more than 4 5 of the weight of the large egg.
2. Standard eggs are sold in packs of 12 while large eggs are sold in packs of 5. Find the probability that the weight of a pack of standard eggs differs from twice the weight of a pack of large eggs by at most 5 grams.

My approach to the problem

1. We have: $$\frac{4}{5}$$ weight of a large egg = $$\frac{4}{5}\cdot65=52g$$
As standard and large eggs are normally distributed, we can calculate the probability that the weight of the standard egg is more than $$\frac{4}{5}$$ the weight of the large egg by:
$$P(X>52) = P (Z>\frac{52-50}{\sqrt{4^2+3^2}}) = P(Z>\frac{2}{5}) = P (Z > 0.4) = 1 - 0.65542 = 0.33458$$ or 33.5%
2. We determine the distribution for one pack of standard eggs by: $$12X = N(12\cdot50,12\cdot3^2) = N(600,108)$$
We determine the distribution for two packs of large eggs by: $$2\cdot5Y = N(2\cdot5\cdot65,2\cdot5\cdot4^2)=N(650,160)$$
We determine the distribution in the weight of the two pack groups: $$10Y - 12X = N(650-600,160+108) = N(50,268)$$
For the weight of a pack of standard eggs differs from twice the weight of a pack of large eggs by at most 5 grams, we need to calculate the probability so that: $$-5\le10Y-12X\le5$$
For $$10Y - 12X \ge -5: P (Z\ge\frac{-5-50}{\sqrt{268}}) = P(Z\ge-3.36) = 0.99961$$
For $$10Y - 12X \le 5: P (Z\le\frac{5-50}{\sqrt{268}}) = P(Z\le-2.75) = 1 - 0.99702 = 0.00298$$
As such, $$P(-3.36\le Z\le-2.75) = 0.99961-0.00298 = 0.99663$$ or 99.6%

Was my solutions to the questions correct? Also, is it acceptable to consider 2 packs of 5 large eggs as 10 large egges? I would really appreciate some comments on my work, and I thank you all for your time and contribution!

• The distribution of twice the weight of one pack of five eggs is not the same as the weight of two packs of five eggs.
– whuber
Mar 14 at 20:08
• I don't follow the meaning of "is more than 4 5 of the weight" in your question. It looks like a copy-paste from an assignment in need of an edit. Do you mean $\frac45$? Or something else? Mar 14 at 21:45

For the first question, the weight of the standard egg (SE) shouldn't be compared with a fixed value ($$52=65\times 4/5$$), since the weight of picked large egg (LE) each time is not fixed but is a random value. Denote the weights of the SE and LE by X and Y respectively. The probability you are looking for should be
$$P\left(X>\frac{4}{5}Y\right)=\int_0^\infty f_Y(y)\left(\int_\frac{4y}{5}^\infty f_X(x)\ dx\right)\ dy.$$
• Re "not really normal:" I'm willing to live in a world where the chance that a chicken egg has a negative mass is less than $10^{-57}$!