What is the probability that in 120 tosses of a fair coin between 40% and 60% (inclusively) will be heads?

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- May 23rd 2006, 07:25 PMswoopesjr01fair coin probability
What is the probability that in 120 tosses of a fair coin between 40% and 60% (inclusively) will be heads?

- May 23rd 2006, 07:37 PMThePerfectHackerQuote:

Originally Posted by**swoopesjr01**

40% of 120 is 48

60% of 120 is 72

Which means the probability is,

$\displaystyle P(48)+P(49)+...+P(71)+P(72)=\sum^{72}_{k=48}P(k)$

But, because of binomial probability we have,

$\displaystyle P(k)={120 \choose k}(1/2)^k(1/2)^{120-k}={120 \choose k}(1/2)^{120}$

Thus,

$\displaystyle \sum^{72}_{k=48}{120 \choose k}(1/2)^{120}\approx .9779$ - May 23rd 2006, 08:35 PMCaptainBlackQuote:

Originally Posted by**swoopesjr01**

The mean number of heads is $\displaystyle 60$, the standard deviation is

$\displaystyle \sqrt{120\times 0.5\times 0.5}=\sqrt{30}$

Between 40% and 60% of 120 is between 48 and 72. To allow for the

fact that we can only have an integer number of heads we work the

normal approximation (which is continuous) for between 47.5 and 72,5

heads.

The z-scores for the ends of the range are:

$\displaystyle

z_{lo}=\frac{47.5-60}{\sqrt{30}}\approx -2.282

$

$\displaystyle

z_{hi}=\frac{72.5-60}{\sqrt{30}}\approx 2.282

$

Now looking the probability of a standard normal random variable having

a value in the range $\displaystyle [-2.28,2.28]$ we find the probability is

$\displaystyle \approx 0.9775$.

This last answer may be compared with PH's calculation based on using

the exact binomial formula.

RonL - May 24th 2006, 01:15 PMThePerfectHacker
Man, I really wish I learned the normal curve theory I look like a noob when I answer these probability questions.

- May 24th 2006, 02:59 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

-Dan