Problems about Normal Distribution

Quote:

Originally Posted by csy0961743

Hey guys, come across this question in text book with just numerical answer, I can't think it through, hope to get help here.

Question

A plaque beside an elevator reads " Capacity 950 pounds, 6 persons". Assume the weight of people entering the elevator are independently and identically distributed with mean=150pounds and standard deviation=30pounds.

1) A person is carrying a suitcase that weighs exactly five times his own weights. What is the probability that the weight limit is exceeded?

2)If there are six people in the elevator, what is the probability that the weight limit is exceeded?

My Answers

1) Let his weight be X~N(150, 30^2)
then 5X+X=950,
X=158.3333 which is the maximum weight of this weir dude who carry a
ridiculously heavy suitcase without being accidentally squashed like a bug.

Prob of exceeding limit = P( X> 158.3333) = P(Z > (158.3333-150)/30 )
= 0.5 - 0.2778 = 0.3XXX

2) What should I do in this part? Everyone can deviate a lot from 150, and sure they don't all weight 158.3333......

For question (1) you need to use two things:

1. If X ~ Normal$\displaystyle (\mu, \sigma^2)$ then aX ~ Normal$\displaystyle (a \mu, a^2 \sigma^2)$.

2. If X ~ Normal$\displaystyle (\mu_X, \sigma_X^2)$ and Y ~ Normal$\displaystyle (\mu_Y, \sigma_Y^2)$ then X + Y ~ Normal$\displaystyle (\mu_X + \mu_Y, \sigma_x^2 +\sigma^2_Y)$.

Now let Y = 5X and define W = X + Y. Calculate Pr(W > 950).

For question (2) you need to define $\displaystyle W = X_1 + X_2 + .... + X_6$ (either generalise 2. above or use Google to find the formula for the sum of n normal variates) and then calculate Pr(W > 950).