Question

Find the probability that 200 tosses of a coin will result in (a) between 80 and 120 heads inclusive, (b) less than 90 heads, (c) less than 85 or more than 115 heads, (d) exactly 100 heads.

Answer 1

a)
`P(80\leq X \leq 120)=0.9957`

b)
`P(X < 90)=0.0793`

c)
`P(X < 85)UP(X>115)=0.034`

d)
`P(X = 100)=0`

Explanation:

1) Notation and data given

n= 200 represent the total tosses

p(head)=p(tails)=1/2=0.5 if is a fair coin

The experiment on this case is tossing 200 times a coin

We can calculate np=200*0.5=100>10 and nq=n(1-p)=200*(1-0.5)=100>10

So then since np>10 and nq>10 we can use the approximation normal to the binomial distribution.

Let X our random variable who represents "the number of heads obtained in 200 tosses from a fair coin". This random variable X follows a normal distribution. And since we have all the conditions satisfied we can calculate the mean and the deviation for the normal distribution

`\mu=np=200*0.5=100`

`\sigma=√(npq)=√(200*0.5*(1-0.5))=5√(2)`

Since X follows a normal distribution we can standarize on this way

`z=(X-\mu)/(√(npq))`

And z is distributed normal with mean= and deviation =1.

This z score would be useful in order to calculate the probabilities required.

2) Part a

`P(80\leq X \leq 120)=P((80-100)/(5√(2))\leq z \leq (120-100)/(5√(2)))=P(-2.83 \leq z \leq 2.83)`

Using properties from the normal distribution we have this

`P(-2.83 \leq z \leq 2.83)=P(z\leq 2.83)-P(Z\leq -2.83)=0.998-0.00233=0.9957`

3) Part b

`P(X<90)=P(z < (90-100)/(5√(2)))=P(z<-1.41)`

And using a the normal standard distribution table or excel we find that:

`P(z<-1.41)=0.0793`

4) Part c

Since the events
`P(X<85)` and
`P(X>115)` are independent, so we can find the probability like this
`P(X<85)UP(X>115)=P(X<85)+P(X>115)`

So we can find individually the probabilities like this:

`P(X<85)=P(z < (85-100)/(5√(2)))=P(z<-2.12)=0.017`

`P(X115)=P(z > (115-100)/(5√(2)))=P(z>2.12)=0.017`

So then:

`P(X<85)UP(X>115)=P(X<85)+P(X>115)=0.017+0.017=0.034`

5) Part d

If we use the normal approximation since the area below the curve for a point is not defined. Then the probability P(X=100) would be 0.