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Good evening everyone, I am attempting a self-study question on probability and am faced with one which really quizzed me.

A study found that a family typically spends between \$500 and \$4500 a month on expenses. Suppose the money spent is uniformly distributed between these amounts. If we randomly select 10 families, what is the probability that at least two families spend more than \$3000 a month on expenses.

My approach to this question is that I present

x = families spending > \$3000 a month

I then proceed on to form a normal distribution as follows:

enter image description here

I was thinking of using the Hypergeomatric distribution for this, but also thought that the normal distribution made sense. I am just puzzled how to move on.

From what I understand, the normal distribution never touches the X-axis, so that will mean that it will be hard for me assert that 500 - 4500 is the range, and thus I was skeptical to convert it to the standard normal.

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Can you explain why you are using a normal distribution? – Glen_b Apr 27 '14 at 5:59
Question says: Suppose the money spent is uniformly distributed between these amounts – wolfies Apr 27 '14 at 6:03
Yeps, wolfies echo my thoughts – user1275515 Apr 27 '14 at 6:12
Please explain your thoughts, user1275515. I suspect you misunderstand @wolfies point. How do you get from "uniformly distributed" to using a normal distribution? – Glen_b Apr 27 '14 at 7:22
A uniform distribution is one which has every value between its maximum and minimum being equiprobable, not at all like a normal distribution, which is more probable near the middle, less probable in the tails and isn't bounded. – Glen_b Apr 27 '14 at 16:12
up vote 2 down vote accepted

The random variables $\{X_i\}$, which are defined as the amount of money family $i$ spends in a given month is known to be uniformly distributed, that is, $X_i \sim \text{U}(500, 4500)$.

For a given sample of size 10, you are asked to compute the probability that at least 2 of them will spend more that \$3000. This is given by $$ \begin{align} &\sum_{j=2}^{10} {10\choose j}\mathbb{P}[X> 3000]^j \mathbb{P}[X\leq 3000]^{10-j}\\ =&\sum_{j=2}^{10} {10\choose j}\left(1-\mathbb{P}[X\leq 3000]\right)^j \mathbb{P}[X\leq 3000]^{10-j}\\ \end{align} $$ using the fact that the spending of the families are distributed independently of each other.

Further, for uniformly distributed random variables, $X\sim \text{U}[l, u]$, we know that $$ \mathbb{P}[X\leq x] = \dfrac{x-l}{u-l} $$

Then we can write $$ \begin{align} &\sum_{j=2}^{10} {10\choose j}\left(1-\mathbb{P}[X\leq 3000]\right)^j \mathbb{P}[X\leq 3000]^{10-j} \\ =&\sum_{j=2}^{10} {10\choose j}\left(\frac{2000}{4500}\right)^j\left(\frac{2500}{4500}\right)^{10-j} \end{align} $$

This can be simplified to $$ =1 - \left({10 \choose 1}\left(\frac{2000}{4500}\right)^1\left(\frac{2500}{4500}\right)^9 + \left(\frac{2500}{4500}\right)^{10}\right) $$

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Please read the self-study guidelines on answering homework questions. I think this goes far beyond helpful hints and strays into doing someone's homework for them, robbing them of the opportunity to actually learn the material. – Glen_b Apr 27 '14 at 7:24
Thanks for the clarification fg & glen. I have not learnt to use the X~U(). I was wondering if this is a question that I can use the PDF of con;tinuous uniform random variable as a starting point? f(x) = 1 / (b-a)? Thanks. – user1275515 Apr 27 '14 at 9:06
Thanks guys. I had been looking at fg's guidance several times and finally made out how this question works out in my notes. Thank you so much. – user1275515 Apr 27 '14 at 9:13
@Glen_b Sometimes all students are looking for is a structure for writing down solutions. The details are incidental. – tchakravarty Apr 27 '14 at 9:14
Thanks fg. I was wondering if there is a typo above? P[ X1 > 3000 and X2 > 3000]? Just another clarification, am I right to say that irregardless of the sign-direction (whether more than or less than), the result will still be the same? Say for example, if if was P(X1<3000) instead, will it still be 2500/4000? – user1275515 Apr 27 '14 at 9:30

Neither the normal distribution nor the hypergeometric distribution apply to this question. There is no underlying assumption of normality here: the probability distribution for the expenses for a single family selected at random is given to be uniform between 500 and 4500. Furthermore, the hypergeometric distribution is not applicable because the outcome of whether a family's expenses exceeds 3000 is independent of any other families sampled among the group of 10. The only two distributions you need for this question are uniform and binomial.

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Thanks everyone. I've attached my working with the corrected numbers. Now I understand the rationale behind the workings above. :)

enter image description here

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Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.

The most important thing to arrive at the correct answer is to realize that you should be using the Binomial Distribution. How to see this?

The binomial distribution gives the probability of k successes in n independent yes/no experiments. In your case, the 'independent yes/no experiments' are the drawing of 10 families. That is, from a large population you randomly (i.e. the draws are independent) pick 10 families and look at their monthly expenses. One such draw is an experiment, and you make 10 of them.

Now, using the same naming conventions, we say that a family which expenses exceed \$3000 is a 'success' (remember, this is just a naming convention to be consistent with the language of the wikipedia article;)). The probability of 'success', denoted p in the article, is thus the probability that a uniform random variable on $[500,4500]$ is larger than 3000.

By reading about the binomial distribution you now know how to find the probability of finding exactly k families with expenses over $3000 in a random sample of 10 families. But the question asks about the probability of at least 2 such families out of 10. So, what do do? Note that the probability of at least 2 such families is the same as the probability of exactly 2 out of 10, plus the probability of exactly 3 out of 10, and so on all the way up to exactly 10 out of 10 such families.

However, you can also note that the complementary event of finding at least 2 such families is finding 0 or 1 such families. That is, if you do not find 2 or more such families in your 10 sampled, you must have found 0 or 1. So you can find the probability you seek by writing $$P(\text{At least 2 out of 10 families spend more than \$3000})=1 - P(\text{0 or 1 families spend more than \$3000})=1-P(\text{No family spends more than \$3000})-P(\text{1 family spends more than \$3000})$$Finally, you see that everything in this last row can be computed using the binomial distribution. I leave the actual computations to you.

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