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Mathematics: Post your doubts here!

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Hey guys, I'm having some trouble with this A2 statistics question. I can't quite figure it out, please help :{

The heights of a population of male students are distributed normally with mean 178cm and standard deviation 5cm. The heights of a population of female students are distributed normally with mean 168cm and standard deviation 4cm. Find the probability that a randomly chosen female is taller than a randomly chosen male.

Let X be the random variable denoting the height of a male student. Then X~N( 178, 5^2 =25)

Let Y be the random variable denoting the height of a female student. Then Y~N( 168, 4^2 =16)

Also, based on the above, we have Y-X~N( -10, 25+16=41)

P( Y>X) =P(Y-X>0)=0.0592 (shown)

Hope this helps. Peace.
 
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express cos^2(2x) in terms of cos4x..... pls help

This simply requires a tweaking of the double angel formula, where cos 2x = 2 cos^2 (x) -1
We rewrite things as cos 4x =2 cos^2 (2x) -1 ; making cos^2 (2x) the subject, we therefore have

cos^2 (2x) = 1/2* (cos4x +1) (shown)

Hope this helps. Peace.
 
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WSGvR.jpg



Please help me solve the above question. Seems easy.. but I'm confused.
Thanks
 
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Let X be the random variable denoting the height of a male student. Then X~N( 178, 5^2 =25)

Let Y be the random variable denoting the height of a female student. Then Y~N( 168, 4^2 =16)

Also, based on the above, we have Y-X~N( -10, 25+16=41)

P( Y>X) =P(Y-X>0)=0.0592 (shown)

Hope this helps. Peace.



THANK YOU SO MUCH
:)
 
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This simply requires a tweaking of the double angel formula, where cos 2x = 2 cos^2 (x) -1
We rewrite things as cos 4x =2 cos^2 (2x) -1 ; making cos^2 (2x) the subject, we therefore have

cos^2 (2x) = 1/2* (cos4x +1) (shown)

Hope this helps. Peace.
Show that (cosx+3sinx)^2 equals 5-4cos2x+3sin2x.........quick.....thanks in advance
 
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Show that (cosx+3sinx)^2 equals 5-4cos2x+3sin2x.........quick.....thanks in advance


(cosx+3sinx)^2 = cos ^2 (x) + 6 sinx cosx + 9 sin ^2 (x)

= 0.5* [cos(2x)+1] + 3 (2 sinx cosx) + 9 * 0.5* [1-cos(2x)] (use the identities cos (2x) = 2 cos ^2 (x) -1 = 1- 2sin ^2 (x) for transformation)

= 0.5 cos (2x) + 0.5 +3 sin (2x) + 4.5 -4.5 cos(2x)

= 5-4cos2x+3sin2x (shown)

Hope this helps. Peace.
 
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thank
(cosx+3sinx)^2 = cos ^2 (x) + 6 sinx cosx + 9 sin ^2 (x)

= 0.5* [cos(2x)+1] + 3 (2 sinx cosx) + 9 * 0.5* [1-cos(2x)] (use the identities cos (2x) = 2 cos ^2 (x) -1 = 1- 2sin ^2 (x) for transformation)

= 0.5 cos (2x) + 0.5 +3 sin (2x) + 4.5 -4.5 cos(2x)

= 5-4cos2x+3sin2x (shown)

Hope this helps. Peace.
thanks a lot
 
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Do we have to memorise all the trigonometric formula etc? or not?
what are specific things to be learned for P2 and P3 ?
 
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∫ (3/2-x)+(4x/4+x^2) dx
help

I am assuming (3/2-x) = 3/(2-x) and (4x/4+x^2) = 4x/ (4+x^2)

Then ∫ (3/2-x)+(4x/4+x^2) dx

= ∫ -(-3/2-x)+2*(2x/4+x^2) dx (shape the both parts of the integral in the format ∫ f '(x)/ f(x) dx, so that this can be reduced to ln |f(x)| )

= -3 ln |2-x| + 2 ln |4+x^2 | +C (shown)

Hope this helps. Peace.
 
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how we can go back from probabillity to its value in terms of phi from normal distribution table?
 
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Find the definite integral of √(5x + 4) from x = 0 to x= 1

This question seems easy but I'm not able to solve it.. please help!
Thanks!
 
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