91Ó°ÊÓ

A pair of balanced dice are rolled 120 times and let X denote the number of rolls on which the sum of two numbers is 7.

The probability of getting sum of two numbers 7 is p=\(\frac{1}{6}\)

Therefore, q=1-p that is\(\frac{5}{6}\)

By the central limit theorem, the distribution of X is approximately normal with mean is

\(\left( {120} \right)\left( {\frac{1}{6}} \right) = 20\)

The standard deviation\(\sqrt {120\left( {\frac{1}{6}} \right)\left( {\frac{5}{6}} \right)} = 4.082\)

Let\(Z = \frac{{\left( {X - 20} \right)}}{{4.082}}\)

Then from the table of the standard normal distribution

\({\rm P}\left( {\left| Z \right| \le 1.96} \right) = 0.95\)

Hence

\(\begin{array}{l}k = \left( {1.96} \right)\left( {4.0282} \right)\\k = 8\end{array}\)

The value of k is 8.