6

I was wondering if I could get some help on this problem.

Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads.

Thanks!

Note: on the study guide I did "1-binomcdf(100,0.50,59)" and still got it wrong

Also, this table is included. enter image description here

This is what it looks like. It seems I got it right, but my professor does not approve of my methods. Note the crossed out "-1/2" and "OK" enter image description here

EDIT: Okay, so the answer is 0.0287. I got that by... 1-P(X* < 9.5/5) =1-P(X* < 1.9) Then look at the table to find 1.9. =1-(0.9713) =0.0287

Steve Kass
  • 14,881
Garrett Simon
  • 61
  • 1
    You could use the normal approximation, using that table. But you should be able to do 1 - binomcdf(100, 0.5, 59). What is your answer? What does the solution say? – nomen Dec 12 '13 at 04:57
  • Ah, it looks like he wanted you to use the normal approximation instead of the long calculation. The long calculation gives better results, so I support your effort, but it does require electronics to perform fast. I'll post an answer. – nomen Dec 12 '13 at 05:05
  • Just the usual rule of thumb in such a case... ;-) – Julien Dec 13 '13 at 01:37

4 Answers4

4

For a solution that you can do off the top of your head, consider that for a sampling process with N draws and probability of success P, the standard deviation of the approximating normal distribution is:

$$\sigma = \sqrt{( N * P * (1-P) )}$$

for $N = 100$ and $P = 0.5, \sigma = 5$

Fortuitously, your question of 60 or more heads corresponds precisely to a z value of $2 * \sigma$ or more, which every stats student should know is about $2.5%$. For greater precision, look it up on the table.

I can't speak for your professor, but if you were my student I would award 110% credit and the offer of a paid assistantship if you presented a quick, accurate answer that did not rely on the use of electronic computing.

Trajan
  • 5,194
  • 2
  • 27
  • 71
Robert L Bell
  • 41
1

At least $60$ is the counterpart of less than $60$. As we're dealing with an integer-valued random variable, less than $60$ is the same as less than or equal to $59$.

So $P(X \geq 60) = 1 - P(X \leq 59)$.

So we use the binomcdf(n,p,k) command, where $n$ is the number of trials, $p$ is the probability of success on a trial, and $k$ is the number of events.

So $P(X) \geq 60 = 1-($ binomcdf(100,0.5,59$)$.

Edit: so it looks like your work is correct. Ask your teacher if they're sure the textbook answer is right.

If you'd like more clarification on the matter, this looks like a good resource.

Newb
  • 17,672
  • 13
  • 67
  • 114
1

It looks like they wanted you to use the normal approximation to the binomial distribution.

Let $X \sim \operatorname{Binomial}_{n,p}$. We can form the "standardized" random variable $X^*$ by taking

$$X^* = \frac{X - \mu}{\sigma} = \frac{X - np}{\sqrt{npq}}$$

As it happens, if $n$ is large and $p$ is "small", then we can approximate $X^*$ by the standard normal random variable $Z$. In particular,

$$ P(X^* < k) \approx P(Z < k)$$

Actually, since $X$ and $X^*$ are discrete, we would probably apply the continuity correction. The point of the continuity correction is to account for the fact that $Z$ is continuous. For example, suppose that $x = 1.3$ is a data point. Presumably, that would be rounded down to $1$ when making a discrete model. The continuity correction captures all the extras that should get rounded down. We include them by doing $P(X^* < 1.5)$.

Okay, so how does the whole problem look?

$$ 1 - P(X < 59) = 1 - P\left(X^* < \frac{59 - (100)*0.5}{\sqrt{100*0.5*0.5}}\right)\\ = 1 - P\left(X^* < \frac{9.5}{25}\right)\\ \approx 1 - P\left(Z < 0.38\right) \approx 1 - P\left(Z < 0.5\right)$$

The last step is the continuity correction.

You can look up the probability that $Z < 0.5$ in that table. Apparently, I made some kind of arithmetic mistake, but that's the gist of it. Try to "debug" it!

Actually, I think the problem is that I made the continuity correction at the wrong time. You want to do it when standardizing, which would look like:

$$ 1 - P(X < 59) = 1 - P\left(X^* < \frac{59.5 - (100)*0.5}{\sqrt{100*0.5*0.5}}\right)$$

nomen
  • 2,707
  • I think I see the error. In the denominator you found 1000.50.5 but did not squareroot it. – Garrett Simon Dec 12 '13 at 05:33
  • Super! I hate it when I make a mistake in an answer and end up editing it a bunch trying to fix it. :) – nomen Dec 12 '13 at 05:34
  • Thank you so much for your help, you are a saint and a scholar. – Garrett Simon Dec 12 '13 at 05:50
  • 1
    In addition to the overlooked square root, you are calculating the wrong probability. The question says "at least 60 heads, which for the continuous version is $P(X\ge 59.5)=1-P(X<59.5)$. You computed $1-P(X^*<58.5)$, which is off by one. – Steve Kass Dec 13 '13 at 02:04
1

There are a lot of confusing answers here! As far as I can tell, your only mistake was to get the wrong value for $binomcdf(100,.50,59)$. It's not $0.284439664$. The correct value is $~0.971556$, which would have given you an answer of $0.028444$ to the question.

Steve Kass
  • 14,881