Chapter 7, Problem 39E

a.

To determine

Obtain the probability that an individual in the age group of 75 years and older who spends less than 15 minutes per day using the computer for leisure.

Obtain the probability that an individual in the age group of 15 to 19 years who spends less than 15 minutes per day using the computer for leisure.

Answer to Problem 39E

The probability that an individual in the age group of 75 years and older who spends less than 15 minutes per day using the computer for leisure is 0.5654.

The probability that an individual in the age group of 15 to 19 who spends less than 15 minutes per day using the computer for leisure is 0.2212.

Explanation of Solution

The exponential parameter value for the age group of 75 years and older is as follows:

$\lambda =\frac{1}{18}\text{minutes}$.

The probability that an individual in the age group of 75 years and older who spends less than 15 minutes per day using the computer for leisure is calculated as follows:

$\begin{array}{c}P\left(\text{Time}\text{spent}<15\right)=1-e^{\frac{-1}{18}\left(15\right)}\\ =1-e^{-\left(0.8333\right)}\\ =1-0.4346\\ =0.5654\end{array}$

Therefore, the probability that an individual in the age group of 75 years and older who spends less than 15 minutes per day using the computer for leisure is 0.5654.

The exponential parameter value for the age group of 15 to 19 years is as follows:

$\lambda =\frac{1}{60}\text{minutes}$.

The probability that an individual in the age group of 15 to 19 years who spends less than 15 minutes per day using the computer for leisure is calculated as follows:

$\begin{array}{c}P\left(\text{Time}\text{spent}<15\right)=1-e^{\frac{-1}{60}\left(15\right)}\\ =1-e^{-\left(0.25\right)}\\ =1-0.7788\\ =0.2212\end{array}$

Therefore, the probability that an individual in the age group of 15 to 19 years who spends less than 15 minutes per day using the computer for leisure is 0.2212.      

b.

To determine

Obtain the probability that an individual in the age group of 75 years and older who spends more than 2 hours.

Obtain the probability that an individual in the age group of 15 to 19 years who spends more than 2 hours.

Answer to Problem 39E

The probability that an individual in the age group of 75 years and older who spends more than 2 hours is 0.0013.

The probability that an individual in the age group of 15 to 19 years who spends more than 2 hours is 0.1353.

Explanation of Solution

The probability that an individual in the age group of 75 years and older who spends more than 2 hours is calculated as follows:

$\begin{array}{c}P\left(\text{Time}\text{spent}>2\text{hours}\right)=e^{\frac{-1}{18}\left(120\right)}\\ =e^{-\left(6.6667\right)}\\ =0.0013\end{array}$

Therefore, the probability that an individual in the age group of 75 years and older who spends more than 2 hours is 0.0013.

The probability that an individual in the age group of 15 to 19 years who spends more than 2 hours is calculated as follows:

$\begin{array}{c}P\left(\text{Time}\text{spent}>2\text{hours}\right)=e^{\frac{-1}{60}\left(120\right)}\\ =e^{-\left(2\right)}\\ =0.1353\end{array}$

Therefore, the probability that an individual in the age group of 15 to 19 years who spends more than 2 hours is 0.1353.      

c.

To determine

Obtain the probability that an individual in the age group of 75 years and older who spends between 30 minutes and 90 minutes using the computer for leisure.

Obtain the probability that an individual in the age group of 15 to 19 years who spends between 30 minutes and 90 minutes using the computer for leisure.

Answer to Problem 39E

The probability that an individual in the age group of 75 years and older who spends between 30 minutes and 90 minutes using the computer for leisure is 0.1822.

The probability that an individual in the age group of 15 to 19 years who spends between 30 minutes and 90 minutes using the computer for leisure is 0.3834.

Explanation of Solution

The probability that an individual in the age group of 75 years and older who spends between 30 minutes and 90 minutes using the computer for leisure is calculated as follows:

  $\begin{array}{c}P\left(\begin{array}{l}\text{Spends between 30 }\\ \text{minutes and 90 }\\ \text{minutes}\end{array}\right)=e^{\frac{-1}{18}\left(30\right)}-e^{\frac{-1}{18}\left(90\right)}\\ =0.1889-0.0067\\ =0.1822\end{array}$     

Therefore, the probability that an individual in the age group of 75 years and older who spends between 30 minutes and 90 minutes using the computer for leisure is 0.1822.

The probability that an individual in the age group of 15 to 19 years who spends between 30 minutes and 90 minutes using the computer for leisure is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Spends between 30 }\\ \text{minutes and 90 }\\ \text{minutes}\end{array}\right)=e^{\frac{-1}{60}\left(30\right)}-e^{\frac{-1}{60}\left(90\right)}\\ =0.6065-0.2231\\ =0.3834\end{array}$

Therefore, the probability that an individual in the age group of 15 to 19 who spends between 30 minutes and 90 minutes using the computer for leisure is 0.3834.

d.

To determine

Obtain the 20^th percentile of the age group of 75 years and older.

Obtain the amount of time, more than which, that eighty percent of the individuals of the age group 75 years and older spend using the computer for leisure.

Obtain the 20^th percentile of the age group of 15 to 19 years.

Obtain the amount of time, more than which, that eighty percent of the individuals of the age group of 15 to 19 years spend using the computer for leisure.

Answer to Problem 39E

The 20^th percentile of the age group of 75 years and older is 4 minutes.

About eighty percent of the individuals of the age group 75 years and older spend more than 4 minutes using the computer for leisure.

The 20^th percentile of the age group of 15 to 19 years is 13.4 minutes.

About eighty percent of the individuals of the age group of 15 to 19 years spend more than 13.4 minutes using the computer for leisure.

Explanation of Solution

Let K be the 20^th percentile value of the age group of 75 years and older.

$\begin{array}{c}1-e^{\frac{-k}{18}}=0.20\\ -\frac{k}{18}=\ln \left(0.80\right)\\ k=-18\ln \left(0.80\right)\\ \cong 4\end{array}$

Therefore, the 20^th percentile of the age group of 75 years and older is 4 minutes.

The remaining 80% of the data is above the 20^th percentile value.

Therefore, eighty percent of the age group of 75 years and older spend more than 4 minutes.

Let S be the 20^th percentile value of the age group of 15 to 19 years.

$\begin{array}{c}1-e^{\frac{-S}{60}}=0.20\\ -\frac{S}{60}=\ln \left(0.80\right)\\ S=-60\ln \left(0.80\right)\\ \cong 13.4\end{array}$

Therefore, the 20^th percentile of the age group of 15 to 19 years is 13.4 minutes.

The remaining 80% of the data is above the 20^th percentile value.

Therefore, eighty percent of the age group of 15 to 19 years spend more than 13.4 minutes.