Question

The mean time taken to learn the basics of a word processor software by a group of new users is 150 minutes with a standard deviation of about 16 minutes. According to chebyshev’s theorem. What is the minimum percentage of students that took between 102 and 198 minutes to learn the basics of the word processor? A. 88.9%B.50%C.75.%D.99%

Answer 1

We are given that distribution has a mean of 150 minutes with a standard deviation of 16 minutes.

According to Chebyshev's theorem, we have that the minimum percentage of students between the given interval is given by:

$1-\frac{1}{k^2}$

Where the value of "k" is how many standard deviations is the interval from the mean:

Where "sd" is the standard deviation.

Now, we can use the given value of the means and the standard deviation to determine the value of "k". Using the upper value of the interval we have:

$150+k(16)=198$

Now, we solve for "k". First, we subtract 150 from both sides:

$\begin{gathered} k(16)=198-150 \\ k(16)=48 \end{gathered}$

Now, we divide both sides by 16:

$k=\frac{48}{16}=3$

Therefore, the value of "k" is 3. Now, we substitute in the formula:

$1-\frac{1}{3^2}=1-\frac{1}{9}=\frac{8}{9}$

Therefore, 8/9 of the students are in the interval. This means a percentage of:

$\frac{8}{9}\times100=88.9\%$

Therefore, the percentage is 88.9%.