91Ó°ÊÓ

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.

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:

$\mathrm{μ}=n×p$

$\mathrm{σ}=\sqrt{n×p×(1−p)}$

The mean and standard deviation are:

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

localid="1650045569819" $$\begin{gathered} \mathrm{σ}=\sqrt{n×p×(1−p)} \\ =\sqrt{10000\left(0.5\right)×(1−0.5}) \\ =50 \end{gathered}$$

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

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 $0.5.$

Here,

$$\begin{gathered} np=10000(0.5) \\ =5000â‰\yen 5 \end{gathered}$$

$$\begin{gathered} n(1−p)=10000(1−0.5) \\ =5000â‰\yen 5 \end{gathered}$$

As a result, normal approximation could be utilized.

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$0.5.$

The probability is calculated as:

$P(X≤4933\text{or}Xâ‰\yen 5067)=P\left(\frac{x−\mathrm{μ}}{\mathrm{σ}}≤\frac{4933−5000}{500}\mathrm{c}\right.\text{Or}\left.\frac{x−\mathrm{μ}}{\mathrm{σ}}â‰\yen \frac{5063−5000}{500}\right)$

$=P(Z≤−1.34\text{or}Zâ‰\yen 1.34)$

$=2×P(Z<−1.34)$

$=0.1802$