Solution

In this question, you are given a graph of the frequency distribution of 50 integer values and are asked to compare the average (arithmetic mean) with the median of the distribution.

In general, the median of a group of n values, where n is even, is obtained by ordering the values from least to greatest and then calculating the average(arithmetic mean) of the two middle values. So, for the 50 values shown in the graph, the median is the average of the 25th and 26th values, both of which are equal to 5. Therefore, the median of the 50 values is 5.

Once you know that the median of the 50 values is 5, the comparison simplifies to comparing the average of the 50 values with 5. You can make this
comparison without actually calculating the average by noting from the graph that of the 50 values, 11 values are 1 unit above 5,

• 16 values are equal to 5,
  
• 10 values are 1 unit below 5, and
  
• 13 values are more than 1 unit below 5.
  

Since the part of the distribution that is below 5 contains 23 values—13 of which are more than 1 unit below 5—and the part of the distribution that is above 5 contains 11 values—none of which is more than 1 unit above 5—the average (arithmetic mean) of the 50 values must be less than 5. The correct answer is Choice B.
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Correct: B
Sum = 1 * 2 + 2 *4 + 3 * 7 + 4 * 10 + 5 * 16 + 6 * 11 = 217
Mean = Sum / total_number = Sum / 50 = 217 / 50 = 4.34

Median is the number that when we sort the numbers in an ascending order, it places exactly in the middle of sequence. Or if number of values is even, median is mean of two middle values. Here we should find the average of 25th and 26th values. By looking at the graph. We see 25th and 26th numbers have value 5, so the median is 5.

So median is bigger.

** Another way is that by looking in the graph we see 2 numbers having value 1, 4 numbers having 2, …, 10 numbers having 4. We know the middle numbers (25 and 26) are in 5. So median is 5. On the other hand for calculating the mean, while having median, we should be aware whether mean is bigger than median or not. So we should consider numbers more than 5 and numbers less than 5 for calculating mean. There are 11 numbers more than 5 and having the value 6. But in the left it seems there are more numbers having value less than 5. Of course if there were 11 numbers having a big value, we couldn’t deduce from the graph.
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For mean
11 values are greater than 5 and 23 values are less than 5 so it will bring mean less than 5
median will be avg of 25th and 26th term i.e 5+5/2 = 5
so clearly B>A

From the diagram, we can see that the 50 values consist of:
Two 1's, four 2's, seven 3's, ten 4's, sixteen 5s and eleven 6's.

So, the MEAN \(\frac{= (2)(1) + (4)(2) + (7)(4) + (16)(5) + (11)(6)}{50}\)

\(= \frac{2 + 8 + 28 + 80 + 66}{50}\)

\(=\frac{184}{50}\)

\(3.68\)
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Since we have an EVEN number of values (50 in total) the MEDIAN will be the average of the 2 middlemost values.
So, when all 50 values are arranged in ascending order, the two middlemost values will be the 25th and 26th values.

Let's list the values in ascending order:
{1,1, 2,2,2,2, 3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4 ...(we've listed 23 values so far. So, keep going...)

{1,1, 2,2,2,2, 3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4, 5,5,5,5,5,5,5,5,5,5,...}
We can stop here.
The two middlemost values are 5 and 5

So, the MEDIAN \(= \frac{5+5}{2} = 5\)
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We get:
Quantity A: \(3.68\)
Quantity B: \(5\)

Answer: B

Cheers,
Brent


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In case of positive skewness (right tail is long relative to left tail) Mean> median>mode and in case of negative skewness(left tail is long relative to right tail) Mean< median <mode. The give distribution is negative skewed so median is greater than mean.

- Skewness approach
A previous poster mentioned this approach. Here is a video that discusses skewness and mean, median and mode:
https://www.youtube.com/watch?time_cont ... 6N_l3Bu-Mc

. I attached a screenshot from the video that I think is helpful.

-. my thoughts on skewed graph
. the mode is located at the peak of the graph
. the mean is pulled towards the long tail (extreme values tend to pull the mean towards them)
. the median is in between the mode and the mean
. remember these points and then you can compare mean and median for any skewed distribution

. I also attached a picture showing how we can use this approach for this problem.

- Non skewness approach
Here is a formula to quickly find the location of the median value, which I find useful:

median location = \(\frac{n+1}{2}\)

For this problem:

median location = \(\frac{50+1}{2}\) = 25.5th value

Whenever the formula has a decimal result, like above, we know the median is the average of the nearest integer below the decimal number and the nearest higher integer. So, in this case we know the median is the average of 25th and 26th value.

Looking at the chart we can see the 25th and 26th value are both 5, so the average of those two numbers is 5, so the median is 5.

As for the mean, I just used a weighted mean.

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