written 8.3 years ago by
teamques10
★ 66k

•
modified 8.3 years ago

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