Question

Ron figures that it takes an average (mean) of 17 minutes with a standard deviation of 3 minutes to drivefrom home, park the car, and walk to an early morning class. a) One day it took Rod 21 minutes to get to class. How many standard deviations from the average is that? Is the z value negative or positive? Why?

Answer 1

Given:

- The Mean:

`\mu=17min`

- The Standard Deviation:

`\sigma=3min`

a) Given a value "x" that represents the number of minutes it takes Rod to get to class, then the z-value measures the difference in Standard Deviation units that this value "x" moves away from the Mean.

Since:

`x=21min`

You need to calculate the z-value with this formula:

`z=(x-\mu)/(\sigma)`

Substituting values and evaluating, you get:

`z=(21-17)/(3)=(4)/(3)\approx1.333`

b) That z-value is positive because the x-value is greater than the Mean. Otherwise, the z-value would be negative.

c) In this case, you have this z-value:

`z=-2.5`

And you need to find "x".

Use the same formula applied in Part a, but this time you need to solve for "x":

`\begin{gathered} -2.5=(x-17)/(3) \\ \\ (-2.5)(3)=x-17 \\ -7.5+17=x \\ x=9.5 \end{gathered}`

In order to determine if it is possible that Rod makes it to class in that amount of time or less, you need to calculate this probability:

`P(z\leq-2.5)`

Using the Standard Normal Distribution Table, you get:

`P(z\leq-2.5)=0.0062`

Since the probability is very low, it means that it is almost impossible for Rod to make it to class in that amount of time or less.

Hence, the answers are:

a)

`z\approx1.333`

b) It is positive because the x-value is greater than the Mean. Otherwise, the z-value would be negative.

c) - Commuting Time:

`x=9.5min`

- No, he couldn't, because the probability is very low.