Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The relative frequency distribution of the weights of all passengers using an elevator is mound-shaped, with mean $150$pounds and standard deviation $35$pounds. Suppose there are n passengers in the elevator, describe and explain the probability distribution of the total weight of n passengers in the elevator.

So, here's how I describe the distribution:

I let $X$ be the total weight of $n$ passengers.

I think that if $n < 30$, then $X$ follows T-Distribution with $n-1$ degrees of freedom.

If $n \geq 30$, by CLT, then $X$ follows a Normal Distribution.

I am also making an assumption that the weights of all passengers are normally distributed.

However, in the later parts of the same question, it asks to show the probability that the total weight of $12$ passengers in the elevator not more than $2000$ is $0.0495$. The only way to get the probability of $0.0495$ apparently is through a normal distribution. If I follow the way I described the distribution above, since $n = 12 < 30$, it should follow a T-Distribution, which yields the probability of $0.064$. But this is not a number the question wanted me to show.

So, I suspect that the way I am thinking about the distribution is wrong. What is wrong with the way I think about the probability distribution of this scenario?

share|improve this question
    
If you assume that the weights of passengers are normally distributed, you don't need a t-distribution. The sum of normal random variables is still a normal random variable. –  Raskolnikov Apr 23 '12 at 8:25
    
ohh..but why don't I need a t-distribution? I thought one of the assumptions for a t-distribution is that the random variables are normally distributed? –  xenon Apr 23 '12 at 8:33
    
A t-distribution is only needed when you consider a ratio of a normal random variable and a chi-square random variable. Usually, this arises in hypothesis testing or confidence intervals when you don't now the population variance. But the nature of your question is different. First, you only need a sum of normal random variables, which is also normal. Second, I assume the values you give are population values. At least, you didn't state otherwise. –  Raskolnikov Apr 23 '12 at 8:45
    
Oh I see.... I think I am confused by a statement in my text that says that "a random variable $t$ follows T-Distribution if the mean of a random sample of size $n$ is taken from a normal population". Thanks! –  xenon Apr 23 '12 at 9:08

1 Answer 1

up vote 4 down vote accepted

There is an error in your assertion that the mean weight of the sample follows a t-distribution. A t-distribution is usually applicable when you don't know the population variance. In this case because you are given a population variance of $35^2$ , it would be okay to say that your sample means are normally distributed.

If it helps, think of the "sample means" as adding up the random variables and taking their means. Or, for a sample size of $12$, $$\text{If }\;\;\; \bar X= \frac1{12}\sum_{i=1}^{12}X_i\;\;\;\;\;\; \text{where }X_i \sim N (150,35^2)\\\text{then }\bar X\sim N\Big (150,\frac{35^2}{12}\Big)$$ It should make sense that the sum of a random variable that is normally distributed yields yet another normal random variable. When the population variance isn't known, you can see how one may run into a problem.

If you make this amend in your approach, I don't see why your answer shouldn't match with the one which your textbook expects

share|improve this answer
    
Even after this, if you still are having problem with it, then please let me know. I'll try to get back to you, or if you think it is necessary, work out the steps for you. Regards, –  Bidit Acharya Apr 23 '12 at 9:15
    
Thank you so much for your explanation! Your equations explained to me very well. Thanks!! :) –  xenon Apr 23 '12 at 9:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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