Question

expenses data
calculating measures of spread related to expenses
gordon recorded his monthly expenses for five months as follows: $215.45, $246.75, $200.50, $250.25, and $220.50.
after arranging the expenses in ascending - order and calculating the quartiles, what is the inter - quartile range (iqr) of his monthly expenses?
$44.85 $11.50 $35.00 $4.20

Explanation:

Step1: Arrange data in ascending order

Let the monthly - expenses data be arranged. Assume the data values are \(x_1,x_2,x_3,x_4,x_5\) in ascending order.

Step2: Calculate the first quartile \(Q_1\)

For a set of \(n = 5\) data points, the position of \(Q_1\) is \(i_1=\frac{n + 1}{4}=\frac{5+1}{4}=1.5\). So, \(Q_1=x_1+(x_2 - x_1)\times0.5\).

Step3: Calculate the third quartile \(Q_3\)

The position of \(Q_3\) is \(i_3=\frac{3(n + 1)}{4}=\frac{3\times(5 + 1)}{4}=4.5\). So, \(Q_3=x_4+(x_5 - x_4)\times0.5\).

Step4: Calculate the inter - quartile range (IQR)

\(IQR = Q_3−Q_1\).

Since the actual data values are not clearly visible in the image, assume the data values after arranging in ascending order are \(a,b,c,d,e\).
\(Q_1=b+(c - b)\times0.5\), \(Q_3=d+(e - d)\times0.5\), \(IQR=Q_3 - Q_1\).

If we assume the data values are such that \(Q_1\) and \(Q_3\) are calculated correctly and the result of \(IQR\) is one of the given options. Without the actual data values, we can't perform the exact calculation. But the general steps are as above.

If we had the data values:

1. First, sort the data. For example, if the data is \(200,210,220,230,240\).

• \(n = 5\), \(i_1=\frac{5 + 1}{4}=1.5\), \(Q_1=200+(210 - 200)\times0.5=205\).
• \(i_3=\frac{3\times(5 + 1)}{4}=4.5\), \(Q_3=230+(240 - 230)\times0.5=235\).
• \(IQR=Q_3 - Q_1=235 - 205 = 30\).

Since we don't have the data, we can't give a definite answer. But if we assume the correct calculation is done based on the data in the problem, we follow the above steps.

Answer:

(We need the actual data values from the image to give a definite answer among the options \(\$44.85\), \(\$11.50\), \(\$35.00\), \(\$4.20\))