A distribution with unknown mean μ has variance equal to 1.5. Use central theorem to find how large a sample should be taken from the distribution in order that the probability will be at least 0.95 that the sample mean will be within 0.5 of the population mean.

Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 10M

Year: May 2015

If $X_1,X_2,X_3…..X_n$….. be a sequence of independent identically distributed RVs with $E(X_i )=μ$ and Var($X_i$ )=$σ^2$, i=1,2…. and if $S_n=X_1+X_2+⋯.X_n$ then under certain general conditions, $S_n$ follows a normal distribution with mean nμ and variance$ nσ^2$ as n tends to infinity.

Corollary :

If $\bar{X} =\frac{1}n(X_1+X_2+⋯.X_n)$ then $E(\bar{X} )=\frac1{n} nμ=μ$ and $Var(\bar{X })=\frac1{n^2} .nσ^2=\frac{σ^2}n$

$∴ \bar{X }$ follows $N(μ,\fracσ{( \sqrt n )})$ as n→∞

We have $E(X_i )=μ$ and $Var(X_i )$=1.5

If $\bar{X}$ is the sample mean then $Z=\frac{(\bar{X} -μ)}{(σ/\sqrt n)}$ is a Standard Normal Variate as n→∞

Given $ |\bar{X}-μ| = 0.5$

$σ=\sqrt {Variance}=\sqrt{1.5} $

$∴|Z|=|\frac{(\bar{X} -μ)}{(σ/√n)}|=|(0.5\sqrt n)/\sqrt {1.5}|=|0.4082 \sqrt n| --- (1)$

From the table we know P(|Z|) > 0.95 when Z=1.96

$0.4082\sqrt n=1.96$

$\sqrt n=\frac{1.96}{0.4082}=23.05$

Hence n must be atleast 24