Question

Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine

Answer 1

Answer:$P(5.111<X<5.291)=P(\frac{5.111-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5.291-\mu}{\sigma})=P(\frac{5.111-5.201}{0.065}<Z<\frac{5.291-5.201}{0.065})=P(-1.38<z<1.38)$And we can find this probability with this difference:

$P(-1.38<z<1.38)= P(z<1.38)-P(Z<-1.38)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  $P(-1.38<z<1.38)=P(z<1.38)-P(z<-1.38)=0.916-0.084=0.832$And since we select a sample of 280 coins we will expect as the number of accepted coins:

$Accepted= 280*0.832= 232.96 \approx 233$

$Rejected= 280-233=47$

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 weights of a population, we assume that this distribution is normal and for this case the distribution for X is given by:

$X \sim N(5.201,0.065)$  

Where $\mu=5.201$ and $\sigma=0.065$

We can find this probability:

$P(5.111<X<5.291)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(5.111<X<5.291)=P(\frac{5.111-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5.291-\mu}{\sigma})=P(\frac{5.111-5.201}{0.065}<Z<\frac{5.291-5.201}{0.065})=P(-1.38<z<1.38)$And we can find this probability with this difference:

$P(-1.38<z<1.38)= P(z<1.38)-P(Z<-1.38)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1.38<z<1.38)=P(z<1.38)-P(z<-1.38)=0.916-0.084=0.832$And since we select a sample of 280 coins we will expect as the number of accepted coins:

$Accepted= 280*0.832= 232.96 \approx 233$

$Rejected= 280-233=47$