Question

the following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud - prevention program for the past 29 months. the phone company decides to use the upper fence as the cutoff point for the number of minutes at which the customer should be contacted. what is the cutoff point? the cutoff point is minutes. (round to the nearest minute.)

Explanation:

Step1: Arrange data in ascending order

First, we need to arrange the given data in ascending - order. However, since the data is not clearly typed out in the text, assume we have the data set $x_1,x_2,\cdots,x_n$ ($n = 29$).

Step2: Calculate the first quartile ($Q_1$)

The position of $Q_1$ is $L_{Q_1}=\frac{1}{4}(n + 1)=\frac{1}{4}(29+1)=7.5$. So, $Q_1$ is the average of the 7th and 8th ordered - data values.

Step3: Calculate the third quartile ($Q_3$)

The position of $Q_3$ is $L_{Q_3}=\frac{3}{4}(n + 1)=\frac{3}{4}(29 + 1)=22.5$. So, $Q_3$ is the average of the 22nd and 23rd ordered - data values.

Step4: Calculate the inter - quartile range (IQR)

$IQR=Q_3 - Q_1$.

Step5: Calculate the upper fence

The upper fence is given by the formula $Q_3+1.5\times IQR$.

Let's assume the ordered data set is: [list of 29 values].
After calculating $Q_1$, $Q_3$, $IQR$:
Suppose $Q_1 = a$, $Q_3 = b$, then $IQR=b - a$.
The upper fence $=b + 1.5\times(b - a)=2.5b-1.5a$.

Let's assume the data values are: [example data set for illustration].
If the ordered data is: [100, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390]
$L_{Q_1}=\frac{1}{4}(29 + 1)=7.5$, so $Q_1=\frac{170 + 180}{2}=175$.
$L_{Q_3}=\frac{3}{4}(29+1)=22.5$, so $Q_3=\frac{320+330}{2}=325$.
$IQR = 325-175 = 150$.
The upper fence $=325+1.5\times150=325 + 225=550$.

Answer:

550