Binomial distribution of coin tosses

[toothpaste666]

Homework Statement

1. A fair coin is tossed 100 times.
(a) Find an approximate probability of getting at least 60 heads.

(b) Find an approximate probability of getting exactly 60 heads.

The Attempt at a Solution

part b) would be b(60;100,.5)

part a) we would need the table for the cumulative distribution, but this is a practice question for my midterm where I won't have access to the tables. Is there a way I can find this with the formula on the test?

 

[davidmoore63@y]

why don't you try approximating with a normal distribution

 

[RUber]

Will you have access to the normal (z) table? You could approximate this as a normal distribution.
Otherwise, I would look at ways to simplify. Since p=1-p=.5, your terms if you were going to write out the sum of p(60)+p(61) + ... +p(100) would all be 100 C k (.5)^100.
You will see that those terms shrink pretty quickly, so depending on how accurate you need your answer, you might be able to only use some of the largest terms--maybe the first 10 or so.
Last option, you could assume that cumulative prob of 50 is .5 +p(50)/2. Then find p(51 - 59). The remaining part is p(at least 60).

 

[Ray Vickson]

Homework Statement

1. A fair coin is tossed 100 times.
(a) Find an approximate probability of getting at least 60 heads.

(b) Find an approximate probability of getting exactly 60 heads.

The Attempt at a Solution

part b) would be b(60;100,.5)

part a) we would need the table for the cumulative distribution, but this is a practice question for my midterm where I won't have access to the tables. Is there a way I can find this with the formula on the test?

Perhaps the tester is looking for a convenient approximation in symbolic form, without necessarily needing a numerical answer. A normal approximation would be the way to go, and specifying correct "z-values" (maybe without being able to actually compute them) would get nearly full marks. Only you and your instructor can know for sure.

 

[toothpaste666]

I will have access to the normal distribution table. For the normal approximation to the binomial distribution

Z = (X - np)/sqrt(np(1-p)) = (60 - 50)/sqrt(25) = 10/5 = 2

we want
1 - F(2)
looking at the standard normal table for
z = 2.00 we have F(z) = .9772
so the answer is
1 - .9772 = .0228