Question

The distribution of number of heads in 100 flips is a discrete histogram, a binomial distribution. But 100 is a lot of bars on a histogram. If, instead, we pretend that the distribution is Normal, with the mean and standard deviations as calculated in Q20, what is the probability of getting between 40 and 60 heads?

Answer 1

Since the distribution of number of heads in 100 flips is a binomial distribution, it is not practical to create a histogram with 100 bars. However, we can use the Normal approximation to the binomial distribution because n=100 is a large value. We know that the mean of the binomial distribution is μ = np = 100 * 0.5 = 50, and the standard deviation is σ = sqrt(np(1-p)) = sqrt(100 * 0.5 * 0.5) = 5.

To find the probability of getting between 40 and 60 heads, we can use the Normal distribution and standardize the values. Let X be the number of heads in 100 flips, then we can write:

P(40 ≤ X ≤ 60) = P((40 - μ)/σ ≤ (X - μ)/σ ≤ (60 - μ)/σ)

= P(-2 ≤ Z ≤ 2), where Z is a standard Normal random variable

Using a Normal distribution table or calculator, we find that the probability of Z being between -2 and 2 is approximately 0.9544. Therefore,

P(40 ≤ X ≤ 60) = P(-2 ≤ Z ≤ 2) ≈ 0.9544

So the probability of getting between 40 and 60 heads is approximately 0.9544.