Probability Density Function

OP

So is it f(x) = 1?

Reply 3

Original post by zkm1223

So is it f(x) = 1?

No. The function plotted on the graph is the pdf, i.e. f(x), and you need to find its equation.
It is in two parts; there will be one equation from 0 to 2, and another from 2 to 6. They are both straight lines.

It is the area underneath the graph that is equal to "1".

Reply 4

OP

so is it f(x) = 2 and f(x) = 6?

Reply 5

Original post by zkm1223

so is it f(x) = 2 and f(x) = 6?

Your answer should be something like:

f(x) = 1.5 for 0 <= x <= 2, and

f(x) = 4.5 - 2x for 2 <= x <= 6, and

zero otherwise.

The actual numbers I've given are NOT the correct values, but just given to illustrate the sort of answer you are looking for.

Reply 6

OP

I think I understand it better now. I think the first equation is f(x) = y for 0 <= x <= 2 but I don't know how to get the second equation.

Reply 7

type_writer

split the curve into a rectangle and a triangle. calculate the area of both of these objects and add them (note you dont know the height, so call this a variable x,y,z or whatever is your favourite letter). the sum of the two areas equals 1 (this is true for all pdfs). solve to find your variable and so your y axis value.

now you know what the y axis value is, you can create 2 equations (1 for 0<x<2 and another for 2<x<6). note that these are straight lines so will be in the from y=mx+c for the triangle bit and just y=something for the rectangle bit.

Reply 8

OP

If we call the height c, is the equation for the rectangle part:

f(x) = c for 0 <= x <= 2?

but i don't understand how to work out the second equation.

Reply 9

Original post by zkm1223

If we call the height c, is the equation for the rectangle part:

f(x) = c for 0 <= x <= 2?

but i don't understand how to work out the second equation.

If the height is "c", then what's the area of the rectangle part in terms of c.

What's the area of the triangle part?

The sum of the two is equal to "1", so you can set up an equation and solve for c, to get the first part of your answer.

For the second part of your answer, you know two points on that descending line (2,c) and (6,0).

Reply 10

OP

Area of rectangle = 2c

Area of triangle =0.5 x 4 x c = 2c

Therefore:

2c + 2c = 1
c= 0.25

Is this right?

Reply 11

Original post by zkm1223

Area of rectangle = 2c

Area of triangle =0.5 x 4 x c = 2c

Therefore:

2c + 2c = 1
c= 0.25

Is this right?

Yes, that that gives you the value for the first part.

So now for the second part you know that your sloping line passes through (2,0.25) and (6,0).

So whatever method you've learnt for determining the equation of a line....

Reply 12

OP

y - y1 = m (x - x1)

y - 0 = -1/16 (x - 6)

y = -1/16 (x - 6)

Is this right?

Reply 13

Original post by zkm1223

y - y1 = m (x - x1)

y - 0 = -1/16 (x - 6)

y = -1/16 (x - 6)

Is this right?

Yes. You can substitute the two values of x, 2 and 6, into the equation to see if they give the right values for y, as a check.

Reply 14

OP

Ok, thanks for all the help. Also please can you tell me how to write out the final answer.

Reply 15

Original post by zkm1223

Ok, thanks for all the help. Also please can you tell me how to write out the final answer.

Have a look at post #6

Reply 16

OP

OK, thanks. Last question. I am asked to calculate the inter-quartile range. How would I do this?

Reply 17

Original post by zkm1223

OK, thanks. Last question. I am asked to calculate the inter-quartile range. How would I do this?

You need to find the value of x, such that the integral from 0 to x is 0.25 (i.e. the area under the graph from 0 to x), and similarly for 0.75, and then subtract the two values to get the interquartile range.

Reply 18

OP

I have integrated f(x) with limits x and 0 and made it equal 0.25. After integrating it all I was left with a quadratic equation which I got the answers x=2 and x=4.

I have done the same thing but made it equal 0.75, however I can't solve the quadratic to get any values of x. Am I doing the right thing?

Reply 19