Question-40
A random variable \(X\) is said to be distributed as \(\text{Bernoulli}( \theta )\), denoted by \(X\sim \text{Bernoulli}( \theta )\), if
\[ P( X=1) =\theta ,\ P( X=0) =1-\theta \]
for \(0< \theta < 1\). Let \(Y=\sum\limits _{i=1}^{300} X_{i}\), where \(X_{i} \sim \text{Bernoulli}( \theta )\), \(i=1,2,\cdots ,300\) be independent and identically distributed random variables with \(\theta =0.25\). The value of \(P( 60\leqslant Y\leqslant 90)\), after approximation through the Central Limit Theorem, is given by
(Recall that \(\phi ( x) =\frac{1}{\sqrt{2\pi }}\int\limits _{-\infty }^{x} e^{-\frac{t}{2}^{2}} dt\))
Let us first state the central limit theorem. Let \(X_{1} ,\cdots ,X_{n}\) be \(n\) i.i.d random variables, each with mean \(\mu\) and variance \(\sigma ^{2}\). Then the distribution of
\[ \frac{X_{1} +\cdots +X_{n} -n\mu }{\sqrt{n} \sigma } \]
tends to the standard normal as \(n\rightarrow \infty\). If we set \(Z=\cfrac{Y-n\mu }{\sqrt{n} \sigma }\), where \(Y=X_{1} +\cdots +X_{n}\), then we can approximate the distribution of \(Z\) with a standard normal for large \(n\).
In this question, \(Y=X_{1} +\cdots +X_{n}\), \(\mu =\theta\), \(\sigma ^{2} =\theta ( 1-\theta )\). Now:
\[ P( 60\leqslant y\leqslant 90) =P( l\leqslant Z\leqslant r) \]
where \(l\) and \(r\) can be computed by plugging in \(60\) and \(90\) in the equation linking \(Z\) and \(Y\). All that remains is to compute the two ends-points:
\[ \begin{aligned} l=\frac{60-n\mu }{\sqrt{n} \sigma } & =\frac{60-300\times 0.25}{\sqrt{300} \times \sqrt{0.25\times 0.75}}\\ & =-2 \end{aligned} \]
The other end-point is:
\[ \begin{aligned} r=\frac{90-n\mu }{\sqrt{n} \sigma } & =\frac{90-300\times 0.25}{\sqrt{300} \times \sqrt{0.25\times 0.75}}\\ & =2 \end{aligned} \]
We are given the CDF of the standard normal to be \(\phi\). The required approximate probability is therefore \(\phi ( 2) -\phi ( -2)\).