Question

For a fair coin, suppose you toss the coin 100 times.

Answer 1

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

• $\mu = np = 100(0.5) = 50$.

• $\sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5$

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is one subtracted by the p-value of Z when X = 66.5.

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

$Z = \frac{66.5 - 50}{5}$

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1