Chapter 2.1, Problem 31E

To determine

Mean and standard deviation of the lengths of the cab rides in miles.

Answer to Problem 31E

Mean in miles,

  ${\mu }_{\text{mi}\text{.}}=4.6667\text{miles}$

Standard deviation in miles,

  ${\sigma }_{\text{mi}\text{.}}=3.7778\text{miles}$

Explanation of Solution

Given information:

  $\text{Fare}=2.85+2.7\left(\text{miles}\right)$

Mean,

  ${\mu }_{\$}=\$15.45$

Standard deviation,

  ${\text{SD}}_{\$}=\$10.20$

Calculations:

In order to obtain the number of miles from the fare, we are required to decrease the fare by 2.85, and then the result to be divided by 2.7.

Since

  $\text{Fare}=2.85+2.7\left(\text{miles}\right)$

That becomes

  $\text{miles}=\frac{\text{Fare}-2.85}{2.7}$

Mean :

Since mean represents the center of the measurement.

When every data value is decreased by 2.85, then the center of the measurement is also decreased by 2.85.

Similarly,

When every data value is divided by 2.7, then the center of the measurement is also divided by 2.7.

Then

The new mean length in miles is the mean fare decreased by 2.85, and then divided 2.7.

New mean,

  ${\mu }_{\text{mi}\text{.}}=\frac{{\mu }_{\$}-2.85}{2.7}=\frac{15.45-2.85}{2.7}=\frac{12.60}{2.7}\approx 4.6667\text{miles}$

Thus,

The mean length of the cab ride in miles is 4.6667 miles.

Standard deviation :

Since the standard deviation is the measure of variability.

When every data value is decreased by 2.85, then the variability of the distribution will not be affected.

And

When every data value is divided by 2.7, then the variability of the distribution is also divided by 2.7.

Then

The new standard deviation

  ${\sigma }_{\text{mi}\text{.}}=\frac{{\sigma }_{\$}}{2.7}=\frac{10.20}{2.7}\approx 3.7778\text{miles}$

Thus,

The standard deviation of the length of the cab ride in miles is 3.7778 miles.