Question

When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the approximate probability that the number of tails is between 40 and 60. Explain your reasoning step by step

Answer 1

96.42% probability that the number of tails is between 40 and 60.

Explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

100 tosses, so
`n = 100`

Two outcomes, both equally as likely. So
`p = (1)/(2) = 0.5`

So

`E(X) = np = 100*0.5 = 50`

`√(V(X)) = √(np(1-p)) = √(100*0.5*0.5) = 5`

Estimate the approximate probability that the number of tails is between 40 and 60.

Using continuity correction.

`P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)`

This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So

X = 60.5

`Z = (X - \mu)/(\sigma)`

`Z = (60.5 - 50)/(5)`

`Z = 2.1`

`Z = 2.1` has a pvalue of 0.9821

X = 39.5

`Z = (X - \mu)/(\sigma)`

`Z = (39.5 - 50)/(5)`

`Z = -2.1`

`Z = -2.1` has a pvalue of 0.0179

0.9821 - 0.0179 = 0.9642

96.42% probability that the number of tails is between 40 and 60.