Answer to Question #269539 in Statistics and Probability for athii

Question

The supervisor of a Muesli Bar producing factory noted that the weight of each 30g bar is

a normally distributed random variable with a mean of 30.25g and a standard deviation of

0.2g. Find the probability that the mean weight of 5 Muesli bars is less than 30g. Suppose

the distribution of weight of each Muesli bar is non-normal. Under what situation would

the above answer not, change?

Answer 1

a) Let $X=$ the mean weight of $n$ Muesli bars: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=30.25\ g, \sigma=0.2\ g, n=5$

$$P(X<30)=P(Z<\dfrac{30-30.25}{0.2/\sqrt{5}})$$

$$\approx P(Z<-2.795085)\approx0.0025943$$

b) By the Central Limit Theorem the normal approximation for $X$ will generally be good if $n \geq 30.$ If $n < 30,$ the approximation is good only if the population is not too different from a normal distribution (if the population is known to be normal, the sampling distribution of $X$ will follow a normal distribution exactly, no matter how small the size of the samples.

The sampling distribution of $X$ will still be approximately normal with mean $\mu_X=30$ and variance $s^2/n=0.2^2/5,$ provided that the sample size $n$ is large $(n \geq 30).$

$$P(X<30)=P(Z<\dfrac{30-30.25}{s/\sqrt{n}})$$

$$=P(Z<\dfrac{30-30.25}{0.2/\sqrt{5}})\approx0.0025943$$