Question

A waiter believes that his tips from various customers have a slightly right skewed distribution with a mean of 10 dollars and a standard deviation of 2.50 dollars. What is the probability that the average of 35 customers will be more than 13 dollars

Answer 1

Answer:

$P(\bar X >13)$

And we can use the z score given by:

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

And replacing we got:

$P(\bar X >13)=P(Z>\frac{13-10}{\frac{2.5}{\sqrt{35}}}=7.09)$

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

$P(Z>7.09)=1-P(Z<7.09)=1-0.999999 \approx 0$

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 variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,2.5)$  

Where $\mu=10$ and $\sigma=2.5$

We select a sample size n = 35, and we know that the distribution for the sample mean $\bar X$ is given by:

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

We want to find the following probability:

$P(\bar X >13)$

And we can use the z score given by:

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

And replacing we got:

$P(\bar X >13)=P(Z>\frac{13-10}{\frac{2.5}{\sqrt{35}}}=7.09)$

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

$P(Z>7.09)=1-P(Z<7.09)=1-0.999999 \approx 0$