91Ó°ÊÓ

The number of people in a city who watch favourable television programs is 15000.

The total population is 500,000.

The number of people who were contacted at random was 200.

Let X be the random variable for the number of people in 200 who watched the program.

The probability of success in one trial,

\(\begin{array}{c}p = \frac{{15,000}}{{500,000}}\\ = 0.03\end{array}\)

As each event is independent, the variable would follow binomial distribution.

That is,

\(X \sim Bin\left( {n = 200,p = 0.03} \right)\)\(\)

The conditions for normal approximations;

\(\begin{aligned}{}np &= 200\left( {0.03} \right)\\ &= 6\\ > 5\\n\left( {1 - p} \right) &= 200\left( {1 - 0.03} \right)\\ &= 194\\ > 5\end{aligned}\)

Thus, the variable X can be approximated to normal with mean and standard deviation as,

\(\begin{aligned}{}\mu &= np\\ &= 200 \times 0.03\\ &= 6\end{aligned}\)

\(\begin{aligned}{}\sigma &= \sqrt {np\left( {1 - p} \right)} \\ &= \sqrt {200 \times 0.03 \times \left( {1 - 0.03} \right)} \\ &= \sqrt {5.82} \\ &= 2.412\end{aligned}\)

Since, \({\bf{X}} \sim {\bf{Normal}}\left( {{\bf{\mu = 6,\sigma = 2}}{\bf{.412}}} \right)\).

Therefore, the normal approximated variable is\({\bf{Z = }}\frac{{{\bf{X - np}}}}{{{\bf{np}}\left( {{\bf{1 - p}}} \right)}} \sim {\bf{N}}\left( {{\bf{0,1}}} \right)\)

The probability that fewer than four are watching the program,

\(\begin{aligned}{}P\left( {X < 4} \right) &= P\left( {X < 4 - 0.5} \right)\\ &= P\left( {X < 3.5} \right)\end{aligned}\)

The probability is computed as,

\(\begin{aligned}{}P\left( {X < 4} \right) &= P\left( {\frac{{X - 6}}{{2.412}} < - 1.036} \right)\\ &= P\left( {Z < - 1.036} \right)\\ &= 0.1500\end{aligned}\)

Thus, the required probability is 0.1500.