According to a survey of the top 10 employers in a major city, a worker spends an average of 413 minutes a day on the job. Suppose the standard deviation is 26.8 minutes and the time spent is approximately a normal distribution.
-Consider the interval $\mu \pm \mathrm { k } \sigma$
For some population.According to Chebyshev's theorem,what value of k would guarantee this interval would include at least 80% of the measurements in the population?
A)5.0
B)2.2
C)2.5
D)1.6
E)2.0

Question

Quizplus · Accepted Answer

Chebyshev's theorem states that for any distribution, the proportion of results within $k$ standard deviations ($\sigma$) of the mean ($\mu$) is at least $1 - \frac{1}{k^2}$ for $k > 1$. To guarantee at least 80% of the measurements, we set $1 - \frac{1}{k^2} \geq 0.8$, solving for $k$ gives $k \geq \sqrt{\frac{1}{0.2}} = \sqrt{5} \approx 2.236$, which is closest to 2.2.

According to a Survey of the Top 10 Employers in a Major