Question

A fair coin is tossed 20 times and all tosses are independent. (a) What is the true probability of obtaining exactly 10 heads? Compute and compare the probability when approximated by an appropriate normal distribution. (b) Compute approximate probability of obtaining more than 7 heads both and without continuity correction. Compare these approximations to the exact probability found using pmf of binomial distribution.

Answer 1

a)
`P(X=10)=(20C10)(0.5)^(10) (1-0.5)^(20-10)=0.176`

b) Using the pmf from the exact distribution

And we can find this probability like this:

`P(X>7) = 1-P(X\leq 7) = 1-[P(X=0)+P(X=1)+......+P(X=7)]`

And we can use the following excel code to find the probability:

"=1-BINOM.DIST(7,20,0.5,TRUE)"

And we got:
`P(X>7) = 1-P(X\leq 7) = 1-[P(X=0)+P(X=1)+......+P(X=7)] =0.868`

Normal approximation

tex] E(X) = np = 20*0.5= 10[/tex]

`Var(X)= np(1-p) = 20*0.5*(1-0.5) = 5`

`Sd(X)= √(5)= 2.236`

Without continuity correction

`P(X>7) = P(Z>(7-10)/(2.236))=1- P(Z<-1.342)= 0.9102`

With continuity correction

For this case we need to do this:

`P(X>7) = P(X>6.5) = P(Z>(6.5-10)/(2.236)) = 1-P(Z<-1.57) = 0.9418`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest "number of heads in 20 trials", on this case we now that:

`X \sim Binom(n=20, p=0.5)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

For this case we can calculate this probability:

`P(X=10)=(20C10)(0.5)^(10) (1-0.5)^(20-10)=0.176`

Part b

We want this probability
`P(X >7)`

Using the pmf from the exact distribution

And we can find this probability like this:

`P(X>7) = 1-P(X\leq 7) = 1-[P(X=0)+P(X=1)+......+P(X=7)]`

And we can use the following excel code to find the probability:

"=1-BINOM.DIST(7,20,0.5,TRUE)"

And we got:
`P(X>7) = 1-P(X\leq 7) = 1-[P(X=0)+P(X=1)+......+P(X=7)] =0.868`

Normal approximation

If we use the normal approximation we need to find the expected value, variance and deviation like this;

`E(X) = np = 20*0.5= 10`

`Var(X)= np(1-p) = 20*0.5*(1-0.5) = 5`

`Sd(X)= √(5)= 2.236`

And we can use the z score given by:

`z= (x -\mu)/(\sigma)`

Without continuity correction

`P(X>7) = P(Z>(7-10)/(2.236))=1- P(Z<-1.342)= 0.9102`

With continuity correction

For this case we need to do this:

`P(X>7) = P(X>6.5) = P(Z>(6.5-10)/(2.236)) = 1-P(Z<-1.57) = 0.9418`