Probability Theory- Standard Normal Distributions

  • Thread starter dzza
  • Start date
  • #1
14
0

Homework Statement



Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 55 percent of the time. We have one of these coins but do not know whether it is a fair coin or a biased one. In order to ascertain which type of coin we have, we shall perform the following statistical test: We shall toss the coin 1000 times. If the coin lands on heads 525 or more times, then we shall conclude that it is a biased coin, whereas, if it lands heads less than 525 times, then we shall conclude that it is the fair coin. If the coin is actually fair, what is the probability that we shall reach a false conclusion? What would it be if the coin were biased?


Homework Equations



The DeMoivre-Laplace limit theorem

The Attempt at a Solution



Since n is large, we can use the normal approximation to the binomial distribution.

We first assume that the coin is fair, that is p = .5

Our mean is np = 1000*.5 = 500
The variance is np(1-p) = 250, so the standard deviation is sqrt(250).

We would reach a false conclusion if the number of heads we observe is greater than or equal to 525.

[tex]P(S_{n} \geq 525) = 1 - P(S_{n} < 525)[/tex]

We now put into form to apply the DeMoivre-Laplace limit theorem:

[tex]1 -P(\frac{S_{n} - np}{\sqrt{np(1-p)}} < \frac{525-500}{\sqrt{250}})[/tex]

[tex]= 1 - \Phi(1.581)[/tex]

Using the table in the book, I look up the the value of the area beneath the standard normal curve to the left of 1.58. This value is .9429.

Hence 1-.9429 = .0571 is the probability of reaching a false conclusion.

The answer in the back of my book is .0606.

I think my method is sound, but I'm maybe acquiring phi incorrectly?
Rounding 1.581 up to 1.59 still doesn't get me to .0606.

Any ideas? I've run into a similar error in other problems, which makes me think I'm doing something systematically wrong.
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,306
48
I don't have a calculator / table at hand, so I cannot check this. But you are only off by a little bit, so I suspect that you might have a slightly incorrect z-value.

Did you apply a continuity correction? You are using a normal (continuous) distribution for a discrete problem. Usually, you adjust your boundary value (525 in this case) a little bit to compensate for that. If n is small enough, that might just mean the difference between .0571 and 0.0606.
 
  • #3
14
0
Thanks CompuChip,

I had not applied the continuity correction to change the value of 525 to 524.5.
From this I was able to get the answer in the back of the book.

I'm not entirely sure why I'm subtracting 1/2 in this case.
The wiki page for continuity correction only seems to showcase adding the quantity 1/2.
The other examples in my book do this as well, except in the case where one wants to find the probability that x is a specific value A, where one then expands that to mean within the interval A +/- 1/2.
In any case, now that I know where the error lies I'll look into it further and hopefully I'll see why.

Thanks again.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
The number of 'heads', N, is a discrete random variable, taking integer values only. Thus, the
event {N >= 125} is the same as the event {N >= 124.5}, and approximating the latter by the normal gives better accuracy, basically because the discrete
probability P{N=k} (for integer k) is being approximated by the probability that the approximating normal variate X lies between k-1/2 and k+1/2.

RGV
 
Last edited:
  • Like
Likes Izzy

Related Threads on Probability Theory- Standard Normal Distributions

  • Last Post
Replies
2
Views
1K
Replies
2
Views
6K
  • Last Post
Replies
8
Views
2K
Replies
1
Views
6K
Replies
7
Views
295
Replies
4
Views
26K
  • Last Post
Replies
5
Views
2K
Replies
2
Views
13K
Replies
1
Views
2K
Top