Question

use the information to answer the question.
a data set is normally distributed. the mean of the data is 9.2, and the standard deviation is 1.8.
using the 68 - 95 - 99.7 rule, what percentage of the data points are greater than 11? enter the answer in the box.
%

Explanation:

Step1: Calculate the number of standard deviations from the mean

The value is \(11\), mean \(\mu = 9.2\), standard deviation \(\sigma=1.8\).
The number of standard deviations \(z=\frac{11 - 9.2}{1.8}=\frac{1.8}{1.8} = 1\)

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(\mu\pm\sigma\) (i.e., \(9.2 - 1.8=7.4\) to \(9.2+1.8 = 11\))
• The total percentage of data is \(100\%\). The percentage of data within \(\mu\pm\sigma\) is \(68\%\). So the percentage of data outside of \(\mu\pm\sigma\) is \(100\%-68\%=32\%\)
• Since the normal distribution is symmetric, the percentage of data greater than \(\mu+\sigma\) is \(\frac{100\% - 68\%}{2}\)

Step3: Calculate the final percentage

\(\frac{100 - 68}{2}\%=\frac{32}{2}\% = 16\%\)

Answer:

\(16\)