Probability

[rad6210]

A fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3

Not sure how to set this up ... any help would be great! thank you!

 

[Deleted member 4993]

A fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3

Not sure how to set this up ... any help would be great! thank you!

It is a binomial distribution.

What are the parameters of the distrbution - when 'n' is large?

 

[wjm11]

A fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3.

I'm a little slow/late with this response, but...

Since you have 10,000 tosses of a fair coin, it is reasonable to approximate the distribution as a standard normal curve. Are you familiar with the 68-95-99.7 rule, which refers to area under a standard normal (bell-shaped) curve? It says that 68% of the area is within +/- one standard deviation of the mean. Your “2/3” is only slightly less than this, meaning that “m” is only slightly less than one standard deviation.

When one has large values for “n”, use the following formula for (binomial distribution) standard deviation:

s.d. = [(n)(p)(1-p)]^(1/2)