According to the chart shown, which of the following is

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Dummies like myself tend to start questions from the basics

Before starting, you need to know how to calculate the Median value - The Median is the "middle number" (in a sorted list of numbers).

To find the Median number, you need to know how many numbers there are in a set, its' total could either be even or odd.
For example

Set 1 ( 2, 5, 6, 7, 8 ) = 5 numbers (odd), the Median value is the 3rd number, 5.

Set 2 ( 2, 4, 6, 8, 10, 11) = 6 numbers (even)0, the Median value is average of the two middle valudes, e.g. (6+8)/2= 7

In the question, 1990 - 2000, there are 11 (odd) numbers in the chart, you will have to sort their values from the lowest to the highest and count the 6th number.

Year '000
1 1991 180
2 1990 190
3 1992 210
4 1993 270
5 2000 270
6 1994 310
7 1995 350
8 1999 360
9 1997 370
10 1996 380
11 1998 390

The correct answer is C

Thanks

Here's a way of doing this problem, without breaking your head is to number the years from left to right, so for 1990, I used 1. 1992, 2, 1993, 3. .....etc.
Then list the same digits in terms of highest value starting from the highest number on the left so for the highest I chose year 9, to the lowest number ( i chose year 1). Make sure you double check the count, so if it's 11 years, you must have 11 digits, organize then pick your median.

In my sequence of numbers I got 9-8-7-10-6-5-4-3-11-2-1, median number shows 5.

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I'm sorry I feel very foolish for getting this wrong and I'm sure this question will sound just as foolish.

I correctly found which bar was the Median bar on the graph (because I know how to find the median) but then estimated that the value of that bar was 325,000 (D). Am I just supposed to be better at reading graphs to know that it's supposed to be 310? I just don't understand how I'm supposed to know that the median bar on the graph is 310 without any other numerical help besides the 100,000 increments on the right. I would definitely get this wrong on the real exam for the same reason.

Well, 6th tallest bar, is a bit more than 300,000 so approximately 310,000.

325,000 would be 300,000 and 1/4th of the increment and the bar for 1994 is less than that.

I marked it E, because the 6th bar is nowhere close to 300,000, its seems to be close to 350,000, or at best close to 325,000. I still feel that the answer choice C is wrong.

Please re-read the solution. We need the value of 6th tallest bar NOT 6th bar from.

Hey thanks a lot. That was a conceptual error on my front. While finding the median I didn't sort the data in ascending / descending order.

as the graph has 11 values, arrange them in ascending order will result in 6th number being the median value which comes out to be approximately 310000. so correct answer option C

Solution:

An easy way to determine which data value is the median is to take the total number of shipments, add 1, and then divide by 2:

(11 + 1)/2 = 6

Thus, the median will be the 6th shipment. (Remember, we will have to list the data values in ascending order, and the 6th value will be the median.)

We can always use the above method of determining the median when we have an odd number of terms.

Let’s start by ordering the approximate number of shipments from least to greatest with the year in parentheses. Since we know the median is the 6th shipment, we can stop once we get to the 6th term.

1) 180,000 (1991)

2) 190,000 (1990)

3) 210,000 (1992)

4) 280,000 (1993)

5) 280,000 (2000)

6) 305,000 (1994)

We see that 305,000 is the median number of shipments, which is closest to 310,000.

The answer is C.

i think the answer is not EXACT; it is approximate figure. @bunual, the tallest 6th bar did not indicate any specific figure. So, why we consider it as (C) 310,000? why don't we consider it as 325,000 (D)?
Thanks...

This question is all about reading graphs.
Since the median value os 1994's value -> Slightly greater than 300,000 => Hence C

2000-1990=10+1(because inclusive)

so, let's find 6th number in ascending order
Answer is C

JeffTargetTestPrep



Is the approach valid for both even and and odd no in data set?
Also arranging in either order shall not make a difference to median.
Am I correct?

Hi Bunuel

Thanks for all your questions and clarifications.

I wonder this answer , when you say 6th is actully the 95' bar chart which is somewhere 340/325 or so. Also with data points for 11 years, the mid point is 6th, am surprised that everytime I see the chart I see 6th as '95 but others are seeing '94. Help ?

Hello experts,
VeritasKarishma, EMPOWERgmatRichC, ArvindCrackVerbal, AaronPond, GMATinsight
If someone (student) has a little bit eye problem, how s/he make decision between C and D?
Thanks__