Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

thank you for reading this over. I'm having trouble starting this problem...

"Suppose that the distribution of the weight of a prepackaged 1-pound bag of carrots is N($1.18, 0.07^2$) and the distribution of the weight of a prepackaged 3-pound bag of carrots is N($3.22, 0.09^2$). Selecting bags at random, find the probability that the sum of three 1-pound bags exceeds the weight of one 3-pound bag."

share|cite|improve this question

closed as off-topic by Mike Miller, hardmath, Claude Leibovici, Graham Kemp, heropup Jun 10 '14 at 5:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, hardmath, Claude Leibovici, Graham Kemp, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.

Let $X_i \sim N(1.18, 0.07^2),\ i = 1, 2, 3$ by the weights of the three prepackaged 1-pound bags, and let $Y \sim N(3.22, 0.09^2)$ by the weight of the prepackaged 3-pound bag. How do you phrase your question in terms of these? – M. Vinay Jun 10 '14 at 2:19
Please show your efforts to solve this problem, @Seraphim. You will receive better responses if it doesn't look like you're just asking people to do your homework for you. – Graham Kemp Jun 10 '14 at 4:32
up vote 5 down vote accepted

Hint: Let $X_1,X_2,X_3$ be the weights of the three one pound bags, and let $Y$ be the weight of the three pound bag.

Let $W=X_1+X_2+X_3-Y$. We want the probability that $W\gt 0$.

Assume independence. Then $W$ has normal distribution, with mean $1.18+1.18+1.18- 3.22$, and variance $(0.07)^2+(0.07)^2+(0.07)^2+ (0.09)^2$.

share|cite|improve this answer
Sorry, I did not read your post before answering; I was just trying to review my Prob&Stats. What are the odds (ha-ha)? – user99680 Jun 10 '14 at 2:27
+1 nothing but quality from you, as always – Asimov Jun 10 '14 at 2:35
@Asimov: thanks for complimenting my jokes :). – user99680 Jun 10 '14 at 2:41
@user99680: No problem. If you want the probability, which you probably don't, it is about $0.98$. – André Nicolas Jun 10 '14 at 2:41

If the random variables $X_1, X_2, X_3, Y$ are each normal and independent, then their sum ( or linear combination ) is also normally-distributed, with mean equal to the sum of their means and variance equal to the sum of the respective variances, e.g. Then construct a new variable $Z$ from these and compute the probability that $Z \geq 0$,.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.