Question

A survey reveals that each customer spends an average of 35 minutes with a standard deviation of 10 minutes in a department store. Assuming the distribution is normal, what is the probability a customer spends less than 30 minutes in the department store?

Answer 1

Answer:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

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 time spent for each customer of a population, and for this case we know the distribution for X is given by:

$X \sim N(35,10)$  

Where $\mu=35$ and $\sigma=10$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(Z$

And the excel code for this case would be : "=NORM.DIST(-0.5,0,1,TRUE)"