Question

5. A coin is tossed 400 times. Use the normal curve approximation to find the probability of obtaining

(a) between 185 and 210 heads inclusive;

(b) exactly 205 heads;

(c) fewer than 176 or more than 227 heads

Answer 1

The probability of getting a head when a coin is tossed 1 time: $p=0.5.$

Let's find the mean:

$\mu=np=400\cdot0.5=200.$

The standard deviation:

$\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot 0.5 \cdot (1-0.5)}=\sqrt{100}=10.$

(a) The probability of obtaining between 185 and 210 heads inclusive:

$P(185\leq X \leq 210)=P(184.5 < X < 210.5).$

Let's find z-scores of 184.5 and 210.5:

$z_1=\frac{184.5-200}{10}=\frac{-15.5}{10}=-1.55,$

$z_2=\frac{210.5-200}{10}=\frac{10.5}{10}=1.05.$

Now we have to use z-table.

$P(184.5 < X < 210.5)=P(-1.55<z<1.05)=0.8531-0.0606=0.7925.$

(b) The probability of obtaining exactly 205 heads:

$P(X=205)=P(204.5<x<205.5).$

Let's find z-scores of 204.5 and 205.5:

$z_1=\frac{204.5-200}{10}=\frac{4.5}{10}=0.45,$

$z_2=\frac{205.5-200}{10}=\frac{5.5}{10}=0.55.$

Now we have to use z-table.

$P(204.5<x<205.5)=P(0.45<z<0.55)=0.7088-0.6736=0.0352.$

(c) The probability of obtaining fewer than 176 or more than 227 heads:

$P(X<176 \: or \: x>227)=P(X<175.5 \: or \: x>227.5)=$

$=P(X<175.5)+P(x>227.5).$

Let's find z-scores of 175.5 and 227.5:

$z_1=\frac{175.5-200}{10}=\frac{-24.5}{10}=-2.45,$

$z_2=\frac{227.5-200}{10}=\frac{27.5}{10}=2.75.$

Then we have:

$P(X<175.5)+P(x>227.5)=P(z<-2.45)+P(z>2.75)=$

$=0.0071+(1-0.9970)=0.0071+0.003=0.0101.$

Answer: (a) 0.7925 (b) 0.0352 (c) 0.0101