Chapter 5, Problem 5.94SE

(a)

To determine

To find the probability distribution for $x$ .

Answer to Problem 5.94SE

The probability distribution for $x$ is binomial distribution.

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=0.4\end{array}$

(b)

To determine

To find the value of $P(X\le 4)$ .

Answer to Problem 5.94SE

The value of $P(X\le 4)$ is $0.2173$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=0.4\end{array}$

Thus, the value of $P(X\le 4)$ can be calculated as,

  $P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

This implies,

  $\begin{array}{}$ P(X=0)=C_0^{15}.{(0.4)}^0.{(1-0.4)}^{15-0}$$ =\frac{15!}{0!(15-0)!}\times {(0.4)}^0\times {(1-0.4)}^{15-0}$$ =0.0005$$ P(X=1)=C_1^{15}.{(0.4)}^1.{(1-0.4)}^{15-1}$$ =\frac{15!}{1!(15-1)!}\times {(0.4)}^1\times {(1-0.4)}^{14}$$ =0.0047$$ P(X=2)=C_2^{15}.{(0.4)}^2.{(1-0.4)}^{15-2}$$ =\frac{15!}{2!(15-2)!}\times {(0.4)}^2\times {(1-0.4)}^{13}$$ =0.0219$$ P(X=3)=C_3^{15}.{(0.4)}^3.{(1-0.4)}^{15-3}$$ =\frac{15!}{3!(15-3)!}\times {(0.4)}^3\times {(1-0.4)}^{12}$$ =0.0634$$ P(X=4)=C_4^{15}.{(0.4)}^4.{(1-0.4)}^{15-4}$$ =\frac{15!}{4!(15-4)!}\times {(0.4)}^4\times {(1-0.4)}^{11}$$ =0.1268$\end{array}$

(c)

To determine

To find the probability that $x\text{ exceeds }5$ .

Answer to Problem 5.94SE

The probability that $x\text{ exceeds }5$ is $0.5968$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=0.4\end{array}$

Thus,the probability that $x\text{ exceeds }5$ can be calculated as,

  $P(X>5)=1-P(X\le 5)$

Thus, $P(X\le 5)$ can be calculated as,

  $P(X\le 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$

For which the values of these $P(X=0),P(X=1),P(X=2),P(X=3),P(X=4)$ is calculated in part (b).

This implies,

  $\begin{array}{l}$ P(X=5)=C_5^{15}.{(0.4)}^5.{(1-0.4)}^{15-5}$$ =\frac{15!}{5!(15-5)!}\times {(0.4)}^5\times {(1-0.4)}^{10}$=0.1859\end{array}$

Thus, we have finally that,

  $\begin{array}{l}P(X\le 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\ =0.0005+0.0047+0.0219+0.0634+0.1268+0.1859\\ =0.4032\end{array}$

  $\begin{array}{l}P(X>5)=1-P(X\le 5)\\ \Rightarrow P(X>5)=1-0.4032\\ \Rightarrow P(X>5)=0.5968\end{array}$

(d)

To determine

To find the largest value of $c$ for which $P(X\le c)\le 0.5$ .

Answer to Problem 5.94SE

  $5$ is the largest value of $c$ for which $P(X\le c)\le 0.5$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=0.4\end{array}$

We have from part (c) that $P(X\le 5)=0.4032$ .

Now, we will find the probability of $P(X\le 6)$ , i.e.

  $P(X\le 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)$

For which the values of $P(X=0),P(X=1),P(X=2),P(X=3),P(X=4),P(X=5)$ are calculate in part (b) and part (c), thus,

  $\begin{array}{l}$ P(X=6)=C_6^{15}.{(0.4)}^6.{(1-0.4)}^{15-6}$$ =\frac{15!}{6!(15-6)!}\times {(0.4)}^6\times {(1-0.4)}^9$=0.2066\end{array}$

Therefore,

  $\begin{array}{l}P(X\le 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)\\ =0.0005+0.0047+0.0219+0.0634+0.1268+0.1859+0.2066\\ =0.6098\end{array}$

Since $P(X\le 5)=0.4032$ and $P(X\le 6)=0.6098$ , $5$ is the largest value of $c$ for which $P(X\le c)\le 0.5$ .