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November 01, 2025, 10:08:26 am
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Author Topic: need help in this question  (Read 1315 times)  Share 

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kriptik

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need help in this question
« on: October 31, 2010, 08:11:28 pm »
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need help
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_avO

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Re: need help in this question
« Reply #1 on: October 31, 2010, 08:14:39 pm »
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Stupid question, you can only determine the range of which the median lies in, but not the actual amount
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sam.utute

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Re: need help in this question
« Reply #2 on: October 31, 2010, 08:16:06 pm »
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What exam was this in again? I remember screwing up this question. I would have guessed the answer as B.
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kriptik

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Re: need help in this question
« Reply #3 on: October 31, 2010, 08:16:48 pm »
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MAV 2010
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shokstar

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Re: need help in this question
« Reply #4 on: October 31, 2010, 08:17:29 pm »
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Well yeh you can only find the range, which is 5-10, and the only alternative that lies within that range is A. It does say what is the BEST alternative, dont see how its a stupid question  :-\
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kriptik

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Re: need help in this question
« Reply #5 on: October 31, 2010, 08:18:28 pm »
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the range, it lies between is 10 and 15 so its either 11.00 or 12.50
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2009
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2010
English [28]   Methods (CAS) [43]   Further maths [50]
Physics [41]   Chemistry        [38]
ATAR: 94.35

2011

shokstar

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Re: need help in this question
« Reply #6 on: October 31, 2010, 08:20:29 pm »
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Quote from: shokstar on October 31, 2010, 08:17:29 pm
Well yeh you can only find the range, which is 5-10, and the only alternative that lies within that range is A. It does say what is the BEST alternative, dont see how its a stupid question  :-\

CRAP disregard this, i was thinking mode ><
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shokstar

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Re: need help in this question
« Reply #7 on: October 31, 2010, 08:23:50 pm »
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Is the answer C? they could be thinking the mid point of the range...yeh its a stupid question  :P
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sam.utute

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Re: need help in this question
« Reply #8 on: October 31, 2010, 08:24:41 pm »
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I get it! Woot!

Dodgy question but solved by:
Looking for the 25th value (which is the median)
It lies within 10-15, but we don't know where.
Assume that the progression will be even (that is, it will increase like 10, 10.5, 11, 11.5, 12 etc.)
As we have 23 terms before this, the 25th term will be two above 10.
So the answer is $11.
The suggeste solution follows this method. Still, it's a stupid question.
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_avO

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Re: need help in this question
« Reply #9 on: October 31, 2010, 08:28:34 pm »
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Can't assume anything unless its stated! gotta remember that
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UncleXxx

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Re: need help in this question
« Reply #10 on: October 31, 2010, 08:42:13 pm »
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Here is what i think.
Find the Mean of the data. For interval data, the mean is worked out by finding the MId point between the intervals and multiplying by it's frequency. Then you plus these numbers  together and divide by the total frequency.

The mean score i got is 11.8877
As you can see, the data is positively skewed, so when graphs are positively skewed, the mean is generally higher than the median. So the median has to be less than 11.8877

Also as Kryptic said "the range, it lies between is 10 and 15 so its either 11.00 or 12.50", so that it should be B

(Not saying that this is 100% right, but just trying to come up with another solution.)
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_avO

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Re: need help in this question
« Reply #11 on: October 31, 2010, 08:44:22 pm »
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Wow great thinking UncleXxx you have proved the impossible!
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Blakhitman

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Re: need help in this question
« Reply #12 on: October 31, 2010, 08:54:43 pm »
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Solution to it is attached.

Question 6
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