91Ó°ÊÓ

Let \(X\) be a random variable that denotes the number of prizes.

Number of trials, \(n=190\)

Probability of success i.e., probability of a winner lotto ticket, \(p=1/5=0.2\)

Mean, \(\mu=np=190*0.2=38\)

Standard deviation, \(\sigma=\sqrt{190*0.2(1-0.2)}=5.5136\)

The required probability is given by,

\(P(34\leq x\leq 54)=P\left ( \frac{34-38}{5.5136}\leq \frac{x-\mu}{\sigma}\leq \frac{54-38}{5.5136} \right )\)

\(P(-0.7255\leq z\leq 2.9019)\)

This probability is shown by the shaded region in the graph below. Hence, it can be written as:

\(P(-0.7255\leq z\leq 2.9019)=P(-0.7255\leq z \leq 0)+P(0\leq z \leq 2.9019)\)

\(=P(0\leq z \leq 0.7255)+P(0\leq z \leq 2.9019)\)

\(=0.2659+0.4982\)

\(=0.7641\)

Therefore, the probability that there are somewhere between \(34\) and \(54\) prizes for the lotto tickets is \(0.7641\).

Mean, \(\mu=np=190*0.2=38\)

Standard deviation, \(\sigma=\sqrt{190*0.2(1-0.2)}=5.5136\)

The required probability is given by,

\(P(54\leq x\leq 64)=P\left ( \frac{54-38}{5.5136}\leq \frac{x-\mu}{\sigma}\leq \frac{64-38}{5.5136} \right )\)

\(P(2.9019\leq z\leq 4.7156)\)

This probability can be written as:

\(P(2.9019\leq z\leq 4.7156)=P(0\leq z\leq 4.7156)-P(0\leq z\leq 2.9019)\)

\(=0.4999-0.4982\)

\(=0.0017\)

Therefore, the probability that there are somewhere between \(34\) and \(54\) prizes for the lotto tickets is \(0.0017\).

Mean, \(\mu=np=190*0.2=38\)

Standard deviation, \(\sigma=\sqrt{190*0.2(1-0.2)}=5.5136\)

The required probability is given by,

\((x>64)=P(\frac{x-\mu}{\sigma}>\frac{64-38}{5.5136}\)

\(P=(z>4.7156)\)

This probability can be written as:

\(P(z>4.7156)=0.5-P(0<z<4.7156)\)

\(=0.5-0.4999999\)

\(=0.0000012\)

Therefore, the probability that there are more than \(64\) prizes for the lotto tickets is \(0.0000012\),which is approximately \(0\).