Plzzzz hurry I have a competitive exam nearby
Asked by SSHHII | 3rd Aug, 2009, 10:19: PM
This problem will require trial based on reasoning.
Let's label the 7 observations according to the ascending order as,
Since median is 23, we have x4 = 23, and since range is 5 i.e. x7-x1 = 5;
Now since mode is 25, at the max. x5, x6, x7 can be 25 or at least two observations (let's say x6 and x7) can be 25.
Let's start with the possibility that x5, x6, x7 are all 25.
Then as x7 is 25 x1 has to be 20 since range is 5.
That leaves just 21 and 22 numbers as possible values of x2 and x3,
so the possible values of x2 and x3 that we can try are 21,21 or 21, 22 or 22, 22.
Let's try 21, 22.
Now we can write out all these values for the observations - 20, 21, 22, 23, 25, 25, 25
Let's verify the average = (20 + 21 + 22 + 23 + 25 + 25 + 25)/7 = 23.
Also we can try other combinations that gives the average of 23.23 or something like that, but not exactly 23.
and I hope it's a single correct answer type question. And anyway if you try any other combinations it'll not work.
So the sum of last 3 observations will be 75.
and the sum of first three obeservations will be 63.
Answered by | 4th Aug, 2009, 08:31: PM
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