Question

While he was a prisoner of war during World War II, John Kerrich tossed a coin $10,000$times. He got$5067$heads. If the coin is perfectly balanced, the probability of a head is $0.5$.

(a) Find the mean and the standard deviation of the number of heads in $10,000$tosses, assuming the coin is perfectly balanced.

(b) Explain why the Normal approximation is appropriate for calculating probabilities involving the number of heads in $10,000$tosses.

(c) Is there reason to think that Kerrich’s coin was not balanced? To answer this question, use a Normal distribution to estimate the probability that tossing a balanced coin $10,000$times would give a count of heads at least this far from $5000$(that is, at least$5067$ heads or no more than $4933$ heads

Short answer

(a)Mean=$5000$

Standard deviation=$50$

(b)$np=5000\ge 5$

$n(1-p)=5000\ge 5$

Thus, normal approximation could be used here.
(c) Since the probability is more than localid="1649834146840" $5\%,$there is no reason to suppose Kerrich's coin was not balanced using probability computation.

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, While he was a prisoner of war during World War II, John Kerrich tossed a coin $10,000$times. He got $5067$heads. If the coin is perfectly balanced, the probability of a head is $0.5$. We need to find the mean and the standard deviation of the number of heads in$10,000$tosses.

Part (a) Step 2: Explanation

Given:

Number of trials $\left(n\right)=10000$

Probability of success $\left(p\right)=0.5$

The mean and standard deviation of a binomial distribution can be calculated using the following formula:

$\mu =n\times p$

$\sigma =\sqrt{n\times p\times (1-p)}$

The mean and standard deviation are:

localid="1650045555230" $$\begin{gathered} \mu =n\times p \\ =10000\left(0.5\right) \\ =5000 \end{gathered}$$

localid="1650045569819" $$\begin{gathered} \sigma =\sqrt{n\times p\times (1-p)} \\ =\sqrt{10000\left(0.5\right)\times (1-0.5}) \\ =50 \end{gathered}$$

Thus, the mean and standard deviation are $5000$and $50.$

Part (b) Step 1: Given information

Given in the question that, While he was a prisoner of war during World War II, John Kerrich tossed a coin $10,000$ times. He got $5067$ heads. If the coin is perfectly balanced, the probability of a head is $0.5.$

Part (b) Step 2: Explanation

Here,

$$\begin{gathered} np=10000(0.5) \\ =5000\ge 5 \end{gathered}$$

$$\begin{gathered} n(1-p)=10000(1-0.5) \\ =5000\ge 5 \end{gathered}$$

As a result, normal approximation could be utilized.

Part (c) Step 1:Given information

Given in the question that, While he was a prisoner of war during World War II, John Kerrich tossed a coin $10,000$ times. He got $5067$ heads. If the coin is perfectly balanced, the probability of a head is$0.5.$

Part (c) Step 2: Explanation 

The probability is calculated as:

$P(X\le 4933\text{or}X\ge 5067)=P\left(\frac{x-\mu }{\sigma }\le \frac{4933-5000}{500}\mathrm{c}\right.\text{Or}\left.\frac{x-\mu }{\sigma }\ge \frac{5063-5000}{500}\right)$

$=P(Z\le -1.34\text{or}Z\ge 1.34)$

$=2\times P(Z<-1.34)$

$=0.1802$