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 0.9 that the sample mean will be within 0.5 of the population mean.

**Mumbai University > Electronics and Telecommunication Engineering > Sem 5 > Random Signal Analysis

Marks: 10M

Year: May 2016

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 X ̅ =1($X_1$+$X_2$+⋯.$X_n$)/n, then E(X ̅ )= nμ/n =μ and Var(X ̅ )=1/$n^2$ .$nσ^2$=$σ^2$/n

∴X ̅ follows N(μ,σ/√n) as n→∞

We have E($X_i$ )=μ & Var($X_i$ )=1.5

If X ̅ is the sample mean then Z=(X ̅-μ)/(σ/√n) is a Standard Normal Variate as n→∞

Given |X ̅-μ|=0.5

σ=√Variance=√1.5

∴|Z|=|(X ̅-μ)/(σ/√n)|=|(0.5√n)/√1.5|=|0.4082√n| .......... (1)

From the table we know P(|Z|)>0.9 when Z=1.64

0.4082√n=1.64

√n=1.64/0.4082

n=16.14142

Hence n must be atleast 17