In $10,000$independent tosses of a coin, the coin landed on heads $5800$times. Is it reasonable to assume that the coin is not fair? Explain.

The total number of tosses$\left(n\right)=10,000$

No. of Heads =$5800$

If the coin is fair, then the probability of obtaining head is = $P\left(head\right)=0.5$.

So, $n=10,000$and $p=0.5$

$$\begin{gathered} np(1-p)=10,000(0.5)(1-0.5) \\ =2500 \end{gathered}$$

As, $np(1-p)\ge 10$, so we will approximate binomial distribution with normal distribution to check that the coin is fair or not.

$$\begin{gathered} \mu =np \\ =10,000(0.5) \\ =5,000 \end{gathered}$$

$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ =\sqrt{10000(0.5)(1-0.5)} \\ =\sqrt{2500} \\ =50 \end{gathered}$$

It is given that the coin landed on heads $5800$times.

Now, we will obtain the probability that the number of heads exceed $5800$.

If it is significantly large, then our assumption will be true, else false.

localid="1647253344667" $$\begin{gathered} P\left(x\ge 5799.5\right)=P\left(\frac{X-np}{\sqrt{np(1-p)}}\ge \frac{5000-np}{\sqrt{np(1-p)}}\right) \\ =P\left(\frac{5799.5-5000}{50}\right) \\ =P(Z\ge 15.99) \\ \approx 0.00 \end{gathered}$$

Since, the above probability is $0$so, our assumption is false. Therefore, the coin is unfair or biased.