# Statistics Probability

A survey of adults found that 34% say their favorite sport is professional football. You randomly select 130 adults and ask them if their favorite sport is professional football.

a) Find the probability that at most 66 people say their favorite sport is professional football.

The steps I have taken to solve this include:

$$n=130,\quad p=.34,\quad q=.66$$

$$\text{mean} = np = 44.2;\quad \text{standard deviation} = \sqrt{ npq} = 5.4$$ So $$z ={ 66 - 44.2 \over 5.4} = 4.03.$$

This z-score is too high to use on my chart. I don't know where to go from here.

The next question I have is how I would find the probability that MORE than 31 people say their favorite sport is professional football. I know I am doing something wrong, can anyone help? Thank you!

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Try the probability calculator at this site – Dilip Sarwate Dec 21 '11 at 17:43
In short, if the $z$-score $z$ is "off the chart, then say $P[Z\ge z]=0$ and $P[Z\le z]=1$ (you are approximating after all). And this brings up a pet peeve of mine: problems that ask you find a quantity (implying they want it exactly) that presume you will approximate the answer using the Central Limit Theorem. – David Mitra Dec 21 '11 at 17:51

Let's assume that each adult answers independently. Let $X$ denote an indicator random variable, equal to 1, if an adult names the professional football his favorite sport, and 0 otherwise. We take the survey's result to mean that $\mathbb{P}(X=1) = p = 0.34$, and $\mathbb{P}(X=0) = 1-p = 0.66$. Thus $X$ is a Bernoulli random variable.

Asking the question of 130 adults, and counting the total of those favoring professional football, gives $$N = X_1 + X_2 + \cdots+ X_{130}$$ The sum of independent Bernoulli random variables follows a binomial distribution, with parameters $n=130$ and $p=0.34$.

Thus $$\mathbb{P}(N \leq 66) = \sum_{m=0}^{66} \binom{130}{m} (1-p)^{130-m} p^m = 0.99971$$ and $$\mathbb{P}(N > 31) = \sum_{m=32}^{130} \binom{130}{m} (1-p)^{130-m} p^m = 0.99168$$

Using the central limit theorem approximation to the above probabilities gives a good agreement. Indeed, $\mu = 130 p = 44.2$ and $\sigma = \sqrt{130 p (1-p)} = 5.4$, so $$\mathbb{P}(N \leq 66) \approx \Phi \left( \frac{66 - 44.2}{5.4} \right) = \Phi( 4.036) = 0.99973$$ and

$$\mathbb{P}(N > 31) \approx 1 - \Phi \left( \frac{31 - 44.2}{5.4} \right) = 1- \Phi( -2.444) = 0.992736$$

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You are fine.

Your $z$-score is correct. So approximately, the required probability is $$P[Z\le {66-44.2\over5.4}]\approx 1$$ (any $z$-score "off the chart" would have corresponding probability approximately 1).

But this should come as no surprise. 130 people were polled, and the expected number of those whose favorite sport is football is 44.2. The probability that at most 66 favor football, intuitively, should be close to 1.

For the second part, proceed as you did in the first part. Note that here you want to find $P[Z>{31-44.2\over5.4}]$.

If you wish, you could use the so called "continuity correction" in the above (that is, compute $P[Z\le {65.5-44.2\over5.4}]$ and $P[Z>{31.5-44.2\over5.4}]$ ).

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