Question

#8. Bottles of a certain lubricant are designed to contain 6.0 oz. The amount in the container is actually a normal random variable with mean 6.1 oz and s.d. .08 oz. To check whether the machine doing the filling is actually meeting specs fill volumes of a sample of 4 bottles are measured. Call these Xi, X2, X3, X4; assume that they are independen. The true mean fill volume μ, is estim ated by tak ng (a) Is Y an unbiased estimator of w/ustify your answer sarn No.is biased becse e samp i a se (b) What is the variance of Y? 2.

Answer 1

8)

a)

Here $X_{i}\sim N(\mu=6.1,\sigma^{2}=0.08^{2})$

Now, $E(Y)=E(\frac{X_{1}+2X_{2}+2X_{3}+X_{4}}{6})=\frac{E(X_{1})+2E(X_{2})+2E(X_{3})+E(X_{4})}{6}$

$=\frac{6.1+2(6.1)+2(6.1)+6.1}{6}=6.1\neq 6$

So, $E(Y)\neq \mu$.Thus, Y is not an unbiased estimator of $\mu$.

b)

Now,$V(Y)=V(\frac{X_{1}+2X_{2}+2X_{3}+X_{4}}{6})=\frac{V(X_{1})+2^{2}V(X_{2})+2^{2}V(X_{3})+V(X_{4})}{6^{2}}$

$=\frac{0.8^{2}+2^{2}0.8^{2}+2^{2}0.8^{2}+0.8^{2}}{6^{2}}=0.1778$

{Covariance terms become 0 since the events are independent}