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.

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

[tex]\mu=np=200*0.5=100[/tex]

[tex]\sigma=\sqrt{npq}=\sqrt{200*0.5*(1-0.5)}=5\sqrt{2}[/tex]

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

[tex]z=\frac{X-\mu}{\sqrt{npq}}[/tex]

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

[tex]P(80\leq X \leq 120)=P(\frac{80-100}{5\sqrt{2}}\leq z \leq \frac{120-100}{5\sqrt{2}})=P(-2.83 \leq z \leq 2.83)[/tex]

Using properties from the normal distribution we have this

[tex]P(-2.83 \leq z \leq 2.83)=P(z\leq 2.83)-P(Z\leq -2.83)=0.998-0.00233=0.9957[/tex]

3) Part b

[tex]P(X<90)=P(z < \frac{90-100}{5\sqrt{2}})=P(z<-1.41)[/tex]

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

[tex]P(z<-1.41)=0.0793[/tex]

4) Part c

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

So we can find individually the probabilities like this:

[tex]P(X<85)=P(z < \frac{85-100}{5\sqrt{2}})=P(z<-2.12)=0.017[/tex]

[tex]P(X115)=P(z > \frac{115-100}{5\sqrt{2}})=P(z>2.12)=0.017[/tex]

So then:

[tex]P(X<85)UP(X>115)=P(X<85)+P(X>115)=0.017+0.017=0.034[/tex]

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.