Question

The table below shows the noon-time temperatures (in \(^{\circ}F\)) recorded in a city over one week.

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
Temperature	66	78	75	69	78	77	70

If \(m\) is the median temperature, \(f\) is the temperature that occurs most often (the mode), and \(a\) is the average (arithmetic mean) of the seven temperatures, which of the following gives the correct order of \(m\), \(f\) and \(a\)?

• A. a < m > f
• B. a < m < f
• C. m < a < f
• D. m < f < a

Hint:

Sort the seven values to read off the median and mode, then add them up and divide by 7 for the mean; compare the three numbers.

Answer 1

The Correct Option is B

Step 1: Sort the data.
The seven temperatures are 66, 78, 75, 69, 78, 77, 70. Sorted in ascending order: 66, 69, 70, 75, 77, 78, 78.

Step 2: Read off the median.
With 7 values, the median is the 4th value in the sorted list.
Counting 66, 69, 70, then the 4th value is 75, so \(m = 75\).

Step 3: Read off the mode.
The value 78 appears twice, more than any other value, so \(f = 78\).

Step 4: Compute the mean.
Add all seven values: \(66 + 78 + 75 + 69 + 78 + 77 + 70 = 513\).
\[ a = \frac{513}{7} \approx 73.29 \]

Step 5: Order the three values.
\(a \approx 73.29\), \(m = 75\), \(f = 78\).
Since \(73.29 < 75 < 78\), the order is \(a < m < f\).

Step 6: Why the other options fail.
Option A claims \(m > f\), but \(m = 75\) is less than \(f = 78\), so it is wrong.
Option C claims \(m < a\), but \(m = 75\) is greater than \(a \approx 73.29\), so it is wrong.
Option D reverses the order between \(a\) and \(m\), but \(a\) is actually the smallest of the three, so it is wrong.

Final Answer:
The correct order is \(a < m < f\).
\[ \boxed{a < m < f} \]