Problems involving random variables and variance, expectation, etc

By Hal 2001 (P3046) on Wednesday, December 13, 2000 - 12:50 am :

Hi,

I got stuck on part (b) of this question. Was wondering whether someone could help out.

X is a random variable having probability density function (p.d.f) f where:

f(x) = (1/h) 0 < x < h
f(x) = 0 Otherwise

a) Given that Y = X(h - X), find E(Y)
b) Find also the probability that Y is greater than (3h² /16).

How I solved part (a)
Need to find E(x) first.
So,

E (X) = \int_{0}^{h} x . f (x) dx =

after some tidy up

= (h / 2)

$Var (X) = \int_{0}^{h} x^{2} . f (x) dx - E^{2} (X) =$ after some tidy up $= (h^{2} / 12)$
Right, now we know that, Var(X) = E(X² ) - E² (X), then re-arrange to get, E(X² ) = (h² /3)

So now to answer part (a),
E(Y) = E(Xh-X² ) = h*E(X) - E² (X) = (h² /6)

I am not too sure what to do for part (b), can someone help out? Much appreciated.

Regards
Hal

By Pras Pathmanathan (Pp233) on Wednesday, December 13, 2000 - 01:43 am :

Hi Hal,

You want the probability that X(h-X) > 3h² /16,
ie probability that

hX-X² > 3h² /16

or

X² -hX+3h² /16 < 0

Now factorise the left hand side (ie by solving the quadratic eqn x² -hx+3h² /16=0 to find out that the roots are h/4 and 3h/4 [do you know how to do this?]), so we need the probability that

(X-h/4)(X-3h/4) < 0 (because X² -hX+3h² /16 = (X-h/4)(X-3h/4)

Draw a graph to convince yourself that the quadratic is less then zero only if h/4 < X < 3h/4, so that's what you want the probability of, which is 1/2.

Post back if you want anything explained more.

Pras

By Hal 2001 (P3046) on Wednesday, December 13, 2000 - 10:53 am :

Hi Pras,

Thank you!

I found the roots of X² - hX + 3h² /16 = 0, by implementing the coefficients of a, b, c into the quadratic formulae, is that ok? So I got like you, h/4 and 3h/4.

I see from the graph that the quadratic < 0, when h/4 < X < 3h/4, like you stated. But its not clear to me why you got the prob of 0.5? Though it is where the min of the graph, the coefficient of h, h/2?

Once again, thank you for your help.
Hal

By Hal 2001 (P3046) on Wednesday, December 13, 2000 - 11:17 am :

Hi,

I am stuck on this question. Appreciate if some could help out.

A card is selected at random from a pack of 10 cards containing even numbers 2,4,6,..,20 and the random variable X represents the number on the card.

a) Find P(X> =15)
b) Find the mean and the variance.

My answers
a) P(X> =15) = P(X=16,18,20) = (1/10) + (1/10) + (1/10) = 3/10
b) not sure?

Regards
Hal

By Hal 2001 (P3046) on Wednesday, December 13, 2000 - 11:25 am :

OK,

I think I got how to get the mean, but still thinking about how to get the var(X).

here is my go:

X = {2,4,6,...,20}

\sum_{r = 1}^{n} 2 r = 2 \sum_{r = 1}^{n} r = 2 \times (n + 1) / 2 = n + 1 = 11

(if you let

n = 10

) but still working on the Var(X)

By Hal 2001 (P3046) on Wednesday, December 13, 2000 - 11:35 am :

I think I got how to get the Var(X) as well, can someone check whether its ok?

create random variables, X and R (discrete)
X = {2,4,6,8,..,20}
R = {1,2,3,..,10}

X = 2R, ok so far?

Then,
Var(X) = Var(2R) = 2² Var(R) = 4 x((n+1)(n-1))/12 = 33 (if you let n=10)

Is that ok?

By Pras Pathmanathan (Pp233) on Wednesday, December 13, 2000 - 04:26 pm :

Hi Hal,

Sorry I rushed the end there. We decided that the probability you want is $P (Y > 3 h 2 / 16) = P (h / 4 < X < 3 h / 4)$ .

Now with a general probability density function (pdf), $f (z)$ , for a random variable $Z$ , the probability that $Z$ lies between $a$ and $b$ is

$P (a < Z < b) = \int_{a}^{b} f (z) dz$

by definition of the pdf. (The limits are $a$ and $b$ , BTW).

So the prob. you want is the integral of $f (x) = 1 / h$ between $h / 4$ and $3 h / 4$ , which will turn out to be 1/2.

Alternatively, you can notice that $X$ is distributed uniformly. This is when the pdf is constant everywhere where it is not zero, and basically means that the probability of $X$ taking a certain value is equal for all the possible values it can take. (It is the generalization of the discrete uniform distn, eg the one on {2,4,..20} above). With the uniform distribution the probability of $X$ lying in any interval is simply the 'length' of the interval (in our case $3 h / 4 - h / 4 = h / 2$ ), divided by the length of the range of values $X$ could be in (ie the length of the interval $0 < x < h$ , which is $h - 0 = h$ ). So the probabilty is $(h / 2) / h = 1 / 2$ . As another example:

$P (h / 6 < X < h / 2 or 3 h / 4 < X < h) = [(h / 2 - h / 6) + (h - 3 h / 4)] / h = 7 / 12$ .

Pras

PS - the variance in your second question is correct.

By Hal 2001 (P3046) on Thursday, December 14, 2000 - 04:29 pm :

Thanks Pras!

I can't answer one part of a question. Here goes:

A door-to-door canvasser tries to persuade people to have a certain type of double-glazing installed. The probability that his canvassing at a house is successful is 0.05.

Find the least number of houses that he must canvass in order that the probabilty of his getting at least one success exceeds 0.99.

My attempt:
P(success)> = 0.99, right?
not sure of what do thereafter.

Help appreciated.
Hal

By Hal 2001 (P3046) on Thursday, December 14, 2000 - 05:04 pm :

I think I got it,
can somone check if its correct?

P(X> =1)
0.99 < = 1 - P(X=0)
0.99 < = 1 - ((0.05)⁰ x (0.95)ⁿ )
n < = ln(0.01) / ln(0.95)
hence, n = 90
is that ok?

Regards
Hal

By Pras Pathmanathan (Pp233) on Thursday, December 14, 2000 - 10:47 pm :

Yep, that's correct.

Pras

By Hal 2001 (P3046) on Thursday, December 14, 2000 - 11:41 pm :

Hi,

I am not sure how to tackle this problem without using the tables.

The random variable X~Po(3.5), find a, b, c, d so that.

(a) P(X 'less than equal to' a) = 0.8576
(b) P(X 'greater than' b) = 0.6792
(c) P(X 'less than or equal to' c) > = 0.95
(d) P(X 'greater than' d) < = 0.005

Thanks in advance.
Hal.

By Brad Rodgers (P1930) on Sunday, December 17, 2000 - 12:27 am :

Sorry, but what does X~Po(3.50) mean?

Thanks,

Brad

By Kerwin Hui (Kwkh2) on Sunday, December 17, 2000 - 04:51 am :

Hal,

I am not sure why you would want to do that, but the only method I know of is by trial and error.

For example, for question a), we require:

$e^{- 3.50} (1 + 3.5 + \frac{1}{2} (3.5)^{2} + \dots + (3.5)^{a} / (a!)) = 0.8576$

so working to 3 sig figs: $\sum_{r = 1}^{a} 3 . 5^{r} / (r!) = 28.4$

Now you work: $3 . 5^{2} / 2 = 6.13$ , $3.53 / 6 = 7.14$ , and $3.54 / 24 = 6.25$ , so we have $a \geq 4$ . From here,

$P (X \geq 4) = 0.6688$

$P (X \leq 5) = 0.8576$

so $a = 5$ .

Obviously the work is easier if you can consult tables.

Kerwin

By Hal 2001 (P3046) on Sunday, December 17, 2000 - 02:12 pm :

Thanks Kerwin.

Brad,

X ~ Po (3.5)

is Poisson distribution, with random discrete variable. With

λ = 3.5

. Does that help?

Also, $P (X = r) = e^{- λ} λ^{r} / r!$ where $r = 0$ , 1, 2, ...