Chapter 5.1, Problem 43E

a.

To determine

Find the value of $P\left(2\right)$.

Answer to Problem 43E

The value of $P\left(2\right)$ is 0.10.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the number of defects in a randomly chosen printed circuit board.

From the probability distribution table, the probability at the point $x=2$ is $P\left(2\right)=0.10$.

Thus, the value of $P\left(2\right)$ is 0.10.

b.

To determine

Find the probability value, $P\left(\text{1 or more}\right)$.

Answer to Problem 43E

The probability value, $P\left(\text{1 or more}\right)$ is 0.50.

Explanation of Solution

Calculation:

The required probability is the sum of the probabilities at $x=1,x=2\text{ and }x=3$.

$\begin{array}{c}P\left(\text{1 or more}\right)=P\left(x\ge 1\right)\\ =P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{1 or more}\right)=P\left(x\ge 1\right)\\ =P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)\\ =0.30+0.10+0.10\\ =0.50\end{array}$

Thus, the probability value, $P\left(\text{1 or more}\right)$ is 0.50.

c.

To determine

Find the probability that at least two circuits are defective.

Answer to Problem 43E

The probability that at least two circuits are defective is 0.20.

Explanation of Solution

Calculation:

The probability that at least two circuits are defective is the sum of the probabilities at $x=2\text{ and }x=3$.

$\begin{array}{c}P\left(\text{At least two are defective}\right)=P\left(x\ge 2\right)\\ =P\left(x=2\right)+P\left(x=3\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{At least two are defective}\right)=P\left(x\ge 2\right)\\ =P\left(x=2\right)+P\left(x=3\right)\\ =0.10+0.10\\ =0.20\end{array}$

Thus, the probability that at least two circuits are defective is 0.20.

d.

To determine

Find the probability that no more than two circuits are defective.

Answer to Problem 43E

The probability that no more than two circuits are defective is 0.90.

Explanation of Solution

Calculation:

The probability that no more than two circuits are defective is the sum of the probabilities at $x=0,x=1\text{ and }x=2$.

$\begin{array}{c}P\left(\text{Not more than 2}\right)=P\left(x\le 2\right)\\ =P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{Not more than 2}\right)=P\left(x\le 2\right)\\ =P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)\\ =0.50+0.30+0.10\\ =0.90\end{array}$

Thus, the probability that no more than two circuits are defective is 0.90.

e.

To determine

Find the mean.

Answer to Problem 43E

The mean value is 0.80.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

$\begin{array}{c}E\left(X\right)={\mu }_X\\ ={\displaystyle \sum x\cdot P\left(x\right)}\end{array}$

The mean of the random variable is obtained as given below:

x	P(x)	$x\cdot P\left(x\right)$
0	0.5	0
1	0.3	0.3
2	0.1	0.2
3	0.1	0.3
Total	1.00	0.8

Thus, the mean value is 0.80.

f.

To determine

Find the standard deviation.

Answer to Problem 43E

The standard deviation is 0.98.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

${\sigma }_X^2={\displaystyle \sum \left[{\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\right]}$

Where ${\mu }_X={\displaystyle \sum \left[x\cdot P\left(x\right)\right]}$ represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

x	P(x)	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
0	0.5	–0.8	0.64	0.320
1	0.3	0.2	0.04	0.012
2	0.1	1.2	1.44	0.144
3	0.1	2.2	4.84	0.484
Total	1.0	2.8	6.96	0.960

Therefore,

${\sigma }^2=0.96$

Thus, the variance is 0.96.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{0.96}\\ =0.98\end{array}$

That is, the standard deviation is 0.98.

g.

To determine

Find the probability that the circuit will function if it has only one defect or no defect.

Answer to Problem 43E

The probability that the circuit will function if it has only one defect or no defect is 0.80.

Explanation of Solution

Calculation:

The probability that the circuit will function if it has only one defect or no defect is the sum of the probabilities at $x=0\text{ and }x=1$.

$P\left(\text{circuit will fuction}\right)=P\left(x=0\right)+P\left(x=1\right)$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{circuit will fuction}\right)=P\left(x=0\right)+P\left(x=1\right)\\ =0.50+0.30\\ =0.80\end{array}$

Thus, the probability that the circuit will function if it has only one defect or no defect is 0.80.