In a previous in-class activity, you learned about the role that [latex]p[/latex] plays in the shape of the binomial distribution. The binomial probability distribution is skewed right if [latex]p < 0.5[/latex], symmetric and approximately bell shaped if [latex]p= 0.5[/latex], and skewed left if [latex]p > 0.5[/latex]. Let’s discuss the role that [latex]n[/latex] plays in its shape.

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Question 1
Recall what you learned in Preview Assignment 8.F. What did you discover about the role that [latex]n[/latex] plays in the shape of the binomial distribution?
For a fixed [latex]p[/latex], as the number of trials, [latex]n[/latex], in a binomial experiment increases, the probability distribution of the random variable [latex]X[/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex]np \geq 10 \mbox{ and } n(1-p) \geq 10[/latex], the probability distribution will be approximately symmetric and bell shaped.
In other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex]np[/latex], and the expected number of failures, [latex]n(1-p)[/latex], are both at least 10.
Question 2
Recall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?
In addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.
In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \mbox{ to } X + 0.5[/latex].
Why is this necessary? Consider the current example:
The probability that George successfully makes exactly 270 free throws is [latex]P(X= 270) = 0.0766[/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex]P(X= 270.2)[/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.
Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[/latex] is a continuous random variable.
The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation: