# Statistics Help

I have some questions that I'm not sure of. Any help is appreciated.

1. 48.7% of Americans have brown eyes. A convention has 5000 people in attendance. Find the mean (expected value),the standard deviation, and the variance for the number of brown-eyed people at such gatherings of 5000 people.

2. A recent survey shows that 70% of families regularly schedule family vacations for the month of April. Find the probability that at least 10 out of 15 randomly selected families regularly schedule family vacations for the month of April.

35% of American households subscribe to TIVO. Suppose that 12 American households are randomly chosen. (Answer both parts for full credit.)

a) How many of them would you expect to have TIVO? ANSWER-4 OR 4.2

b) What is the probability that 2 OR 3 of them have TIVO?

1. Type O blood is found in 44% of Americans. Suppose 7 samples of blood are tested at random. Find the probability that exactly 2 of the tested samples will be Type O.

2. Forty percent of Clinton County residents are uncomfortable using the Internet. Suppose 11 such people are randomly selected. What is the probability exactly 5 are uncomfortable using the Internet?

3. A basketball player with a history of making 80% of the foul shots taken during games. (Answer all parts for full credit)

1.What is the probability that he will make exactly 6 of 10 foul shots? 2.What is the probability that he will make at least 6 of 10 foul shots? 3.What is the probability that he will MISS at least one of 10 foul shots?

1. Given the following probability distribution, calculate the mean, variance, and standard deviation.

x P(x)

1 0.125

2 0.236

3 0.543

4 0.096

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What, you won't even show what you've tried doing? –  Ｊ. Ｍ. Sep 29 '11 at 1:51
I'm in the process of doing it now. I was hoping to have the answers to check my work against before I submitted. It's an online course with timed quizzes. –  Thomas Sep 29 '11 at 2:06
so its the solution you seek thats it? –  Bhargav Sep 29 '11 at 2:07
Yes. If I have a different answer I'll post and try to figure out where I went wrong. –  Thomas Sep 29 '11 at 2:11
I have answered the last question, and posted it separately. I would appreciate any feedback –  Thomas Sep 29 '11 at 2:45

We look at the following multipart problem, which sadly cannot be identical to any of the ones that you listed, since I do not wish to be part of the writing of an online test. (By the way, thanks for being straightforward about that.) However, you will find that most of the questions you asked have close analogues below. I hope these will be helpful.

Sixty percent of Lardon County residents are obese.

(a) $12$ Lardon residents are chosen at random. What is the probability that exactly $9$ of them are obese.

Solution: We are not given the population of Lardon County, nor are we told how the sampling was done (with replacement, a person might be chosen more than once, or without replacement). Technically, we assume that it is with replacement, though the population of Lardon is presumably large enough compared to $12$ that whether it is with or without replacement makes no real difference to the answer.

If a chosen person is obese, call that a success. The probability of success with a randomly chosen person is $0.60$ ($60$ percent). So the probability of failure is $0.40$. We want the probability of exactly $9$ successes in $12$ trials. By a basic formula ("binomial distribution") the probability of exactly $9$ successes in $12$ trials is $$\binom{12}{9}(0.60)^9 (0.40)^3.$$

Here by $\binom{12}{9}$ we mean "$12$ choose $9$", the number of ways of choosing $9$ objects from $12$ objects. On your calculator keys, $\binom{n}{r}$ often appears as ${}_nC_r$. Your online notes may call it something else, like $C^n_r$, or $C(n,r)$. By the way, $\binom{12}{9}=220$.

(b) Again, we have $12$ randomly chosen Lardon residents. What is the probability that $9$ or more of them are obese?

Solution: In part (a), we found the probability that exactly $9$ are obese. Similarly, the probability that exactly $10$ are obese is $\binom{12}{10}(0.60)^{10}(0.40)^2$, the probability that exactly $11$ are obese is $\binom{12}{11}(0.60)^{11}(0.40)^1$, and the probability that exactly $12$ are obese is $\binom{12}{12}(0.60)^{12}(0.40)^0$. For the probability $9$ or more are obese, add up. We get $$\binom{12}{9}(0.60)^9 (0.40)^3 +\binom{12}{10}(0.60)^{10}(0.40)^2+\binom{12}{11}(0.60)^{11} (0.40)^1 +\binom{12}{12}(0.60)^{12} (0.40)^0.$$ You may have a calculator or program that evaluates this sort of sum with a few key presses. That part would be device-dependent, and I cannot help with it.

(c) How many people in our sample of $12$ would you expect to be obese?

Solution: If you perform an experiment independently $n$ times, and the probability of success each time is $p$, then the expected number (mean number) of successes is $np$. (This is a fact you are expected to know.) In our case, $n=12$, and the probability of success is $0.60$, so one would "expect" $(12)(0.60)$ people to be obese. Note that this is $7.2$, which is not an integer. That's OK, $7.2$ is the right answer.

(d) In our sample, what is the variance of the number of obese people? What is the standard deviation of the number of obese people?

Solution: If you perform an experiment independently $n$ times, and the probability of success each time is $p$, then the variance of the number of successes is $np(1-p)$. (This is a fact you are expected to know.) so in our case, the variance is $(12)(0.60)(0.40)$. The standard deviation is the square root of the variance. So that you can check you are doing things right, numerically the standard deviation is about $1.697$.

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(+1) ... "sadly" indeed! –  The Chaz 2.0 Sep 29 '11 at 3:33