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.

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.

Accepted Answer

# 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

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Chebyshev's theorem formula

Step2: Solve for $k$

Step3: Convert the mean to minutes

Step4: Calculate the lower and upper bounds

Answer: