Chapter 5.1, Problem 46E

a.

To determine

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

Answer to Problem 46E

The value of $P\left(2\right)$ is 0.258.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the number of teenagers who have sent text messages on their cell phones within past 30 days.

From the probability distribution table, the probability at the point $x=2$ is, $P\left(1=2\right)=0.258$.

Thus, the value of $P\left(2\right)$ is 0.258.

b.

To determine

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

Answer to Problem 46E

The probability value, $P\left(\text{More than 1}\right)$ is 0.888.

Explanation of Solution

Calculation:

The required probability can be obtained as shown below:

$\begin{array}{c}P\left(\text{More than 1}\right)=P\left(x>1\right)\\ =1-P\left(x\le 1\right)\\ =1-\left[P\left(x=0\right)+P\left(x=1\right)\right]\end{array}$

Substituting the value from the probability distribution table,

$\begin{array}{c}P\left(\text{More than 1}\right)=1-\left[P\left(x=0\right)+P\left(x=1\right)\right]\\ =1-\left[0.015+0.097\right]\\ =1-0.112\\ =0.888\end{array}$

Thus, the probability value, $P\left(\text{More than 1}\right)$ is 0.888.

c.

To determine

Find the probability that three or more of the teenagers sent text messages.

Answer to Problem 46E

The probability that three or more of the teenagers sent text messages is 0.63.

Explanation of Solution

Calculation:

The probability that three or more of the teenagers sent text messages is the sum of the probabilities at $x=3,x=4\text{ and }x=5$.

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{Three or more sent text messages}\right)=P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\\ =0.343+0.227+0.06\\ =0.63\end{array}$

Thus, the probability that three or more of the teenagers sent text messages is 0.63.

d.

To determine

Find the probability that fewer than two of the teenagers sent text messages.

Answer to Problem 46E

The probability that fewer than two of the teenagers sent text messages is 0.112.

Explanation of Solution

Calculation:

The probability that fewer than two of the teenagers sent text messages is the sum of the probabilities at $x=3\text{ and }x=4$.

$\begin{array}{c}P\left(\text{Fewer than two sent text messages}\right)=P\left(x<2\right)\\ =P\left(x=0\right)+P\left(x=1\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{Fewer than two sent text messages}\right)=P\left(x<2\right)\\ =P\left(x=0\right)+P\left(x=1\right)\\ =0.015+0.097\\ =0.112\end{array}$

Thus, the probability that fewer than two of the teenagers sent text messages is 0.112.

e.

To determine

Find the mean.

Answer to Problem 46E

The mean value is 2.85.

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	0.015	0.000
1	0.097	0.097
2	0.258	0.516
3	0.343	1.029
4	0.227	0.908
5	0.060	0.300
Total	1.000	2.850

Thus, the mean value is 2.85.

f.

To determine

Find the standard deviation.

Answer to Problem 46E

The standard deviation is 1.107.

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	$P\left(x\right)$	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
0	0.015	–2.85	8.1225	0.122
1	0.097	–1.85	3.4225	0.332
2	0.258	–0.85	0.7225	0.186
3	0.343	0.15	0.0225	0.008
4	0.227	1.15	1.3225	0.300
5	0.060	2.15	4.6225	0.277
Total	1.000	–2.1	18.235	1.226

Therefore,

${\sigma }^2=1.226$

Thus, the variance is 1.226.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{1.226}\\ =1.107\end{array}$

That is, the standard deviation is 1.107.