91Ó°ÊÓ

A fair coin is tossed 900 times.

Hence n=900

For i=1,2,...900

\({X_i} = 1\)if head is obtained on i^thtoss and

\({X_i} = 0\)Otherwise

Therefore\(E\left( {{X_i}} \right) = \frac{1}{2}\)and

\(Var\left( {{X_i}} \right) = \frac{1}{4}\)

It follows from the central limit theorem that the distribution of the total number of heads is\(H = \sum\limits_{i = 1}^{900} {{X_i}} \)will be approximately the normal distribution for which

Mean is\(\left( {900} \right)\left( {\frac{1}{2}} \right) = 450\)and

Variance is\(\left( {900} \right)\left( {\frac{1}{4}} \right) = 225\).

Standard deviation is\(\sqrt {225} = 15\)

Therefore the variable\(Z = \frac{{\left( {H - 450} \right)}}{{15}}\)will have approximately standard normal distribution.

Thus,

\(\begin{array}{c}{\rm P}\left( {H > 495} \right) = {\rm P}\left( {H \ge 495.5} \right)\\ = {\rm P}\left( {Z \ge \frac{{\left( {495.5 - 450} \right)}}{{15}}} \right)\end{array}\)Using continuity correction.

\({\rm P}\left( {H > 495} \right) \approx 1 - \phi \left( {3.033} \right)\)

\({\rm P}\left( {H > 495} \right) \approx 0.0012\).

Using the standard normal distribution table.

HenceProbability of obtaining more than 495 heads is\(0.0012\).