Normal approximation

• Nov 25th 2008, 07:23 PM
someone21
Normal approximation
Two questions
1.From past data, it is known that 70 percent of brand A computer users plan to buy a newer model of brand A computer in future. Suppose 25 brand A computer users are asked for their future brand preference. Use normal approximation to find the chance that exactly 10 of these users intend to switch brand in their next computer purchase.

2.Randomly select a number from the box with numbers -2,-2,-2, 0, 2, 2 and the number obtained is the amount of dollars you get. Play the game 30 times, use normal approximation to find the chance that you will be in the , losing some money.

The point is I dont know how to use normal approximation when finding exactly
• Nov 26th 2008, 02:16 AM
Quote:

The point is I dont know how to use normal approximation when finding exactly
I am glad. It is a completely useless skill.

The exact value is much easier than the approximation: use

$P(X=x) = {n\choose x}p^x(1-p)^{n-x}$

If you really must use the normal approximation, the key is the formula P(X=x) = 1-P(X<x)-P(X>x).
• Nov 26th 2008, 07:28 AM
someone21
Quote:

I am glad. It is a completely useless skill.

The exact value is much easier than the approximation: use

$P(X=x) = {n\choose x}p^x(1-p)^{n-x}$

If you really must use the normal approximation, the key is the formula P(X=x) = 1-P(X<x)-P(X>x).

Yes I have all tried that but seem not to get the correct answer for the first question and have no idea for the second question
• Nov 27th 2008, 03:52 AM
vincisonfire
I think you may use a Poisson law.
$P(k) = \frac{e^{-\lambda}\lambda^k}{k!}$
Here, you expect $25\cdot 0.7=17.5$ user to choose brand A.
Therefore $\lambda=17.5$
$P(10) = \frac{e^{-17.5}17.5^{10}}{10!} = 0.01864 \approx = 1.9\%$
You can try and see that the most probable value is 17 or 18 at about 10%.
Notice that when $\lambda$ grows, your probability distribution looks more and more like a gaussian curve.
• Nov 28th 2008, 01:10 AM
someone21
Quote:

Originally Posted by vincisonfire
I think you may use a Poisson law.
$P(k) = \frac{e^{-\lambda}\lambda^k}{k!}$
Here, you expect $25\cdot 0.7=17.5$ user to choose brand A.
Therefore $\lambda=17.5$
$P(10) = \frac{e^{-17.5}17.5^{10}}{10!} = 0.01864 \approx = 1.9\%$
You can try and see that the most probable value is 17 or 18 at about 10%.
Notice that when $\lambda$ grows, your probability distribution looks more and more like a gaussian curve.

I have used ur way but the answer does not come correct.Moreover I cannot use this law for my course to find the answer
• Nov 28th 2008, 10:46 PM
Quote:

Yes I have all tried that but seem not to get the correct answer for the first question
What exactly are you doing? Do you get 0.0013248 from the binomial distribution and 0.0008849 using the normal approximation?

Quote:

I think you may use a Poisson law.
The Poisson distribution is used to find the probability of numbers of rare events in a set interval. This question requires the binomial distribution, as whether each of the 25 computer users chooses brand A can be considered a Bernoulli distribution and the sum of Bernoulli distributions is binomial.

Quote:

2.Randomly select a number from the box with numbers -2,-2,-2, 0, 2, 2 and the number obtained is the amount of dollars you get. Play the game 30 times, use normal approximation to find the chance that you will be in the , losing some money.
Sorry, forgot about this one. I would suggest using X as the number of wins and Y as the number of losses. Then X has a binomial distribution with parameters (30,2/6) and Y has a binomial distribution with parameters (30-X,3/4). From this you can determine the joint pmf and then figure out how likely it is that Y>X.
• Nov 29th 2008, 01:39 AM
someone21
Quote:

What exactly are you doing? Do you get 0.0013248 from the binomial distribution and 0.0008849 using the normal approximation?

This is my way tell me what is wrong

P(9.5<X<10.5)=P(Z=9.5-25(0.3)/25(0.3)(0.7)SR<X<10.5-7.5/25(0.3)(0.7)SR)

=P(0.87<X<1.31)
=0.0971 but the answer is 0.096

The Poisson distribution is used to find the probability of numbers of rare events in a set interval. This question requires the binomial distribution, as whether each of the 25 computer users chooses brand A can be considered a Bernoulli distribution and the sum of Bernoulli distributions is binomial.

Sorry, forgot about this one. I would suggest using X as the number of wins and Y as the number of losses. Then X has a binomial distribution with parameters (30,2/6) and Y has a binomial distribution with parameters (30-X,3/4). From this you can determine the joint pmf and then figure out how likely it is that Y>X.

Sorry I don't know understand it

• Nov 30th 2008, 03:10 PM
Quote:

P(9.5<X<10.5)=P(Z=9.5-25(0.3)/25(0.3)(0.7)SR<X<10.5-7.5/25(0.3)(0.7)SR)
This indeed is the correct formula, although it is usually written
$P(9.5$

I don't know why you are getting 0.0971 with this. When I enter it on my computer I get 0.09615141.

0.096 is indeed the correct answer. Sorry about giving incorrect ones before; I was using p=0.7 instead of 0.3.

Quote:

I would suggest using X as the number of wins and Y as the number of losses. Then X has a binomial distribution with parameters (30,2/6) and Y has a binomial distribution with parameters (30-X,3/4). From this you can determine the joint pmf and then figure out how likely it is that Y>X.
Remember the formula $P(X=x\cap Y=y) = P(Y=y|X=x)P(X=x)$
• Dec 2nd 2008, 05:08 AM
someone21
Quote:

This indeed is the correct formula, although it is usually written
$P(9.5$

I don't know why you are getting 0.0971 with this. When I enter it on my computer I get 0.09615141.

0.096 is indeed the correct answer. Sorry about giving incorrect ones before; I was using p=0.7 instead of 0.3.

I did was then 0.87<Z<1.31

And what i get is 0.0971 (or maybe all this because i have a paper table which is only 2 dp for Z)

Remember the formula $P(X=x\cap Y=y) = P(Y=y|X=x)P(X=x)$

I think I need to use normal approximation so then.......

Sorry for asking so many questions btw