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: Identify the number of standard deviations from the mean

First, we find how many standard deviations 11 is from the mean. The formula to calculate the number of standard deviations (z - score) is $z=\frac{x - \mu}{\sigma}$, where $x = 11$, $\mu=9.2$ and $\sigma = 1.8$.

Substitute the values: $z=\frac{11 - 9.2}{1.8}=\frac{1.8}{1.8} = 1$.

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 1 standard deviation of the mean ($\mu\pm\sigma$).
• Approximately 95% of the data lies within 2 standard deviations of the mean ($\mu\pm2\sigma$).
• Approximately 99.7% of the data lies within 3 standard deviations of the mean ($\mu\pm3\sigma$).

Since the data is symmetric about the mean, the percentage of data greater than $\mu+\sigma$ (because $z = 1$ means $x=\mu+\sigma$) is calculated as follows:

The total percentage of data is 100%. The percentage of data within $\mu-\sigma$ and $\mu+\sigma$ is 68%, so the percentage of data outside this interval is $100\% - 68\%=32\%$. Since the distribution is symmetric, the percentage of data greater than $\mu+\sigma$ is half of 32%, which is $\frac{32\%}{2}=16\%$.

Answer:

16