You have 3 family birthdays coming up and for each birthday there is a fixed probability p = 0.8 of getting invited to the party, in which case you will be expected to bring a gift. You only have 2 gifts and 5ft of wrapping paper. The length of wrapping paper required for each gift is normally distributed with mean 3ft and standard deviation 1ft. Calculate the probability that you don’t have enough wrapping paper for attend 2 or 3 parties. I got the answer for 2 parties is 0.69 while the right answer is 0.76, the command says the correct answer required using X1 + X2,the sum of lengths of wrapping 2 gifts, but many people wrote down 2X1 instead. I don't know how to do that. Also for 3 parties the answer is 0.99, can anyone help me out?
A question using normal distribution
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statistics
normal-distribution
1 Answers
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In your problem, $X_1 \sim N(3, 1)$ and $X_2\sim N(3, 1)$ (that is, each is distributed as a Normal distribution with mean 3 and variance 1). Assuming $X_1, X_2$ are independent, we know that $X_1+X_2 \sim N(6, 2)$. The probability of not having enough paper is $P(X_1+X_2 > 5)$, and if you look up on any Normal table (or online), you see this probability is indeed 0.760. Note that using $2X_1$ is plain wrong, as $2X_1 \sim N(6, 4)$ instead of $N(6, 2)$.
In the case of three gifts, still assuming independence, $X_1+X_2+X_3 \sim N(9, 3)$, and then any Normal table will tell you that $P(X_1+X_2+X_3>5)=0.989$.