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A2 Statistics

f(x) = x/8 for 0 <= x <= 4
= 0 otherwise

find median value of x

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Original post by uzer
f(x) = x/8 for 0 <= x <= 4
= 0 otherwise

find median value of x

Find F(x)F(x).
Then use the fact that F(m)=12F(m) = \frac{1}{2}.
Reply 2
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uzer
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f(x) is given.. i have done that and then i intergrated a got x^2/16 and then my asnwr is wrong
Original post by uzer
f(x) is given.. i have done that and then i intergrated a got x^2/16 and then my asnwr is wrong

*I, I, integrated, and, answer

Yes, that is correct.
So F(x)=x216F(x) = \frac{x^2}{16}
∴m216=12\therefore \frac{m^2}{16} = \frac{1}{2}
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uzer
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Original post by morgan8002
*I, I, integrated, and, answer

Yes, that is correct.
So F(x)=x216F(x) = \frac{x^2}{16}
∴m216=12\therefore \frac{m^2}{16} = \frac{1}{2}


so then the answer comes out as 4, but my book says 2.8 :frown:
Original post by uzer
so then the answer comes out as 4, but my book says 2.8 :frown:

Show the rest of your workings, you are correct so far and I have m=22=2.83m = 2\sqrt{2} = 2.83(3sf)
Reply 6
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uzer
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m2 = 0.5 times 16

m2 = 8

square root both sides m =4
Original post by uzer
m2 = 0.5 times 16

m2 = 8

square root both sides m =4

8≠4\sqrt{8}\not=4


8=42\sqrt{8} = \sqrt{4}\sqrt{2}
∴8=22=2.83\therefore \sqrt{8} = 2\sqrt{2} = 2.83(3sf)
Reply 8
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uzer
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yes thank you its late at night so
Original post by uzer
yes thank you its late at night so

You're welcome. I thought that might be the case. Maybe in future use a calculator if you're tired.
Reply 10
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uzer
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hi please differentiate

2x/9 (3-x)

for 0<=x<=3

i solved f'(x) for zero and got 0.5.

Please confirm
Original post by uzer
hi please differentiate

2x/9 (3-x)

for 0<=x<=3

i solved f'(x) for zero and got 0.5.

Please confirm

f(x)=2x9(3−x)f(x) = \frac{2x}{9}(3-x)
∴f′(x)=23−4x9\therefore f'(x) = \frac{2}{3}- \frac{4x}{9}

Are you trying to find f'(0) or the value of x for f'(x) = 0
f′(0)=23f'(0) = \frac{2}{3}
f′(x)=0f'(x) = 0 at x=32x= \frac{3}{2}
Reply 12
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Original post by uzer
hi please differentiate

2x/9 (3-x)

for 0<=x<=3

i solved f'(x) for zero and got 0.5.

Please confirm


f(x) = a/x for 1<=x<=2
0 otherwise

find median value of x. i have differentiated but am failing to get the correct answer
Original post by uzer
f(x) = a/x for 1<=x<=2
0 otherwise

find median value of x. i have differentiated but am failing to get the correct answer

Show your working and I can tell you where you went wrong.
Reply 14
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uzer
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f'(X) = -2ax^-2 =0
now i am stuck
Original post by uzer
f'(X) = -2ax^-2 =0
now i am stuck

You integrate to find the median
It should be F(x)=aln⁡(x)F(x) = a\ln(x)



As a side note, the general convention is the pmf is denoted by f(x) and the cdf is denoted by F(x)
(edited 9 years ago)
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Original post by morgan8002
You integrate to find the median
It should be F(x)=aln⁡(x)F(x) = a\ln(x)



As a side note, the general convention is the pmf is denoted by f(x) and the cdf is denoted by F(x)


and then i times by x is that correct
so its x a ln(x) = 0.5
Original post by uzer
and then i times by x is that correct
so its x a ln(x) = 0.5

No I think you're confusing the median with mean.

To find the mean, you do ∫−∞∞xf(x)dx\displaystyle\int_{-\infty}^{\infty} xf(x)dx

To find the median, you find F(x), where F(x)=∫−∞xf(x)dxF(x) = \displaystyle\int_{-\infty}^x f(x)dx, then use F(m)=12F(m) = \frac{1}{2}


For each of these, the infinities just mean the limits of where the function is defined, for your question it will be 1 and 2.

So you should do aln⁡m=12a\ln m = \frac{1}{2}
Reply 18
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uzer
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2^x + 1 divided by 2^x -1 = 5

i tried cross multiplying but no luck
Original post by uzer
2^x + 1 divided by 2^x -1 = 5

i tried cross multiplying but no luck

2x+12x−1=5\frac{2^x + 1}{2^x - 1} = 5
2x+1=5(2x−1)2^x + 1 = 5(2^x -1)
And carry on from there

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