Question

time spent online americans spend an average of 4 hours per day online. if the standard deviation is 26 minutes, find the range in which at least 88.89% of the data will lie. use chebyshevs theorem. round your k to the nearest whole number. at least 88.89% of the data will lie between and minutes.

Explanation:

Step1: Recall Chebyshev's theorem formula

The formula for Chebyshev's theorem is $1-\frac{1}{k^{2}}$, where $1 - \frac{1}{k^{2}}$ is the proportion of data within $k$ standard - deviations of the mean. We are given that $1-\frac{1}{k^{2}}=0.8889$.

Step2: Solve for $k$

\[

$$\begin{align*} 1-\frac{1}{k^{2}}&=0.8889\\ \frac{1}{k^{2}}&=1 - 0.8889\\ \frac{1}{k^{2}}&=0.1111\\ k^{2}&=\frac{1}{0.1111}\approx9\\ k& = 3 \end{align*}$$

\]

Step3: Convert the mean to minutes

The mean is 4 hours. Since 1 hour = 60 minutes, the mean $\mu=4\times60 = 240$ minutes. The standard - deviation $\sigma = 26$ minutes.

Step4: Calculate the lower and upper bounds

The lower bound $L=\mu - k\sigma$ and the upper bound $U=\mu + k\sigma$.
\[

$$\begin{align*} L&=240-3\times26\\ &=240 - 78\\ &=162 \end{align*}$$

\]
\[

$$\begin{align*} U&=240+3\times26\\ &=240 + 78\\ &=318 \end{align*}$$

\]

Answer:

162, 318