frequency distribution

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carred: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sun Jul 03, 2011 9:32 am

frequency distribution

by carred » Tue Jul 05, 2011 7:00 am

Hi

I am stuck at the following example so I could use a little help how to solve it.

Question:
Find the mean and median of the values of the random variable X, whose relative frequency distribution is given in the table below.

X.......Relative Frequency
0.......0.18
1.......0.33
2.......0.10
3.......0.06
4.......0.33

I calculated the mean as follows: 1*0.33+2*0.1+3*0.06+4*0.33 = 2.03

But how is the median calculated? I thought the median of an odd set of numbers is the number in the middle, and the median of an even set of numbers is the mean of the two numbers in the middle.
Therefore the median should be 2 but the solution says it is 1.

Please help :roll:

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Source: Beat The GMAT — Data Sufficiency | Tue Jul 05, 2011 7:00 am

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Tue Jul 05, 2011 8:02 am

carred wrote: Question:
Find the mean and median of the values of the random variable X, whose relative frequency distribution is given in the table below.

X.......Relative Frequency
0.......0.18
1.......0.33
2.......0.10
3.......0.06
4.......0.33

I calculated the mean as follows: 1*0.33+2*0.1+3*0.06+4*0.33 = 2.03

But how is the median calculated? I thought the median of an odd set of numbers is the number in the middle, and the median of an even set of numbers is the mean of the two numbers in the middle.
Therefore the median should be 2 but the solution says it is 1.

I've never seen a real GMAT question which abstractly discussed a 'random variable', or which provided a frequency table that didn't include whole numbers, so this probably isn't an important example to study. The table above is just another way of saying that 18% of the values in your set are '0', 33% of the values are '1', and so on. If x is the median of this set, then it must be true that at least half of the values are less than or equal to x *and* that at least half of the values are greater than or equal to x. So '2' cannot be the median, since only 49% of values are greater than or equal to 2; the median of the set must be 1.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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carred: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sun Jul 03, 2011 9:32 am

by carred » Tue Jul 05, 2011 8:34 am

Alright that sounds obvious. Thanks a lot.

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uppaljayant: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Dec 24, 2011 5:15 am

by uppaljayant » Sat Dec 24, 2011 5:18 am

The median of a discrete random variable is the "middle" value. It is the value of X for which P(X < x)
is greater than or equal to 0.5 and P(X > x) is greater than or equal to 0.5.

Google it out for more details!

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pm; Thanked: 158 times; Followed by:21 members

by pemdas » Sat Dec 24, 2011 7:38 am

carred wrote:Question: Find the mean and median of the values of the random variable X, whose relative frequency distribution is given in the table below.

X.......Relative Frequency
0.......0.18
1.......0.33
2.......0.10
3.......0.06
4.......0.33

I calculated the mean as follows: 1*0.33+2*0.1+3*0.06+4*0.33 = 2.03

hi carred, Ian is absolutely correct in advising about mean as the point for which half is less than or equal and half is more than or equal than the mean multiplied by the number of elements below and/or above mean. This is the so called technically correct mathematical approach for deciding about mean. However, given GMAT's time pacing and "anguishing" environment of taking exam, I doubt even math PhD would apply his soft skills here

instead I would apply hard skills. You correctly calculated the sum for values and now would need to find the expected average (arithmetic mean for freq.distribution). Also, it's advisable to exclude 0 event with 18% probability at all, as this is meaningless for our distribution data; we could have millions of 0 events with various probabilities but their impact on our expected average would be none!

So you need to set (2.03+X*p)/11 where p is the probability (or relative frequency according to your question format). Here's how I would solve this question. I would substitute X with 1 and p with the expected average. But I need to find the expected average first, and this is possible if I do this without X, as X with allowed value of 1 event may be attached our mean anyways. So, 2.03/10=0.203 <- expected average (or mean)
X=1, mean=0.203. What's left is median and we need to order our data accordingly
0.06 0.06 0.06 0.1 0.1 0.203 0.33 0.33 0.33 0.33 0.33
you could avoid ordering the data, as new number of events, 11, implies you set the sixth element (in the ordered row of data) as a median. Our mean is 0.203 and it's also our median.

Mean=Median=0.203

If you doubt about this method and think that adding new element, i.e. 1 event with the assigned frequency (probability) 0.203 would not be sufficient for our decision about median, then you could use any number of new events say 100 and assign them the only valid for the given raw of data frequencies, i.e. expected average equal to 0.203. This would not change our median; it's still 0.203

adding up some points beyond GMAT. If the question would ask me if this is normal distribution? I would answer No, because for meeting normal distribution conditions, the mode must also be equivalent to mean.

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uppaljayant: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Dec 24, 2011 5:15 am

by uppaljayant » Sat Dec 24, 2011 9:16 am

^But the answer is Median = 2!
How do you explain that?

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pm; Thanked: 158 times; Followed by:21 members

by pemdas » Sat Dec 24, 2011 10:35 am

uppaljayant wrote:^But the answer is Median = 2!
How do you explain that?

listen, the solution per original poster says median=1 and not 2!

carred wrote:Therefore the median should be 2 but the solution says it is 1.

Alternatively, you could pick the number of events if such is requested by question or given in answer choices A-E, because GMAT solution always depends on answer choices. So, mean for data containing X of which the assigned probability=the expected average placed right in the middle and is equal to the median of frequency distribution, X is such that it's 1 (or you can say X=1). We decide 1 is the number of events with the probability 0.203 in this data raw, and we mark 1 as the answer.

hope this helps

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