Question

The weight of a package of mints is believed to be normally distributed with mean 21.37 grams and standard deviation 0.4, i.e. X \sim N(21.37,0.4). What is the chance that a sample of 4 packages of mints has an average weight, \overline X, between 21 and 22 grams

Answer 1

Answer:

$P(21<\bar X <22)=P(\frac{21-21.37}{\frac{0.4}{\sqrt{4}}}<Z<\frac{22-21.37}{\frac{0.4}{\sqrt{4}}})$

And using a calculator, excel or the normal standard table we have that:

$P(-1.85<Z<3.15)=P(z<3.15)-P(Z<-1.85) =0.9992-0.03215=0.967$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weigths of a population, and for this case we know the distribution for X is given by:

$X \sim N(21.37,0.4)$  

Where $\mu=21.37$ and $\sigma=0.4$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required with the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(21<\bar X <22)=P(\frac{21-21.37}{\frac{0.4}{\sqrt{4}}}<Z<\frac{22-21.37}{\frac{0.4}{\sqrt{4}}})$

And using a calculator, excel or the normal standard table we have that:

$P(-1.85<Z<3.15)=P(z<3.15)-P(Z<-1.85) =0.9992-0.03215=0.967$