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

Answer:

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

Step-by-step 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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

100 tosses, so $n = 100$

Two outcomes, both equally as likely. So $p = \frac{1}{2} = 0.5$

So

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

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60.5 - 50}{5}$

$Z = 2.1$

$Z = 2.1$ has a pvalue of 0.9821

X = 39.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{39.5 - 50}{5}$

$Z = -2.1$

$Z = -2.1$ 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.