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: Find the mean using the assumed-mean (deviation) method.
Take $A = 75$ as a convenient assumed mean, and find how far each value is from it: $66-75=-9$, $78-75=3$, $75-75=0$, $69-75=-6$, $78-75=3$, $77-75=2$, $70-75=-5$.

Step 2: Add up the deviations.
Sum of deviations $= -9 + 3 + 0 - 6 + 3 + 2 - 5 = -12$.

Step 3: Adjust the assumed mean.
True mean $= A + \frac{\text{sum of deviations}}{n} = 75 + \frac{-12}{7} = 75 - 1.71 = 73.29$. This matches the direct-sum result, but skips adding seven large numbers directly.

Step 4: Find the median and mode by frequency counting.
Listing the values with how many times each appears: 66(1), 69(1), 70(1), 75(1), 77(1), 78(2). Arranged in order (66, 69, 70, 75, 77, 78, 78), the middle (4th) value is 75, so the median $m = 75$. The value with the highest count is 78 (count 2), so the mode $f = 78$.

Step 5: Compare the three figures.
$a = 73.29$, $m = 75$, $f = 78$. Since $73.29 < 75 < 78$, the relation is $a < m < f$.

Final Answer:
The correct ordering is $a < m < f$, matching option B. \[ \boxed{a < m < f} \]