Question

A data set is normally distributed with mean 18 and standard deviation 2.7. Approximately what percent of the data are greater than 14?
A. 0.7%
B. 6.9%
C. 0.9%
D. 93.1%

Answer 1

To find the percentage of the data that is greater than 14 in a normally distributed dataset, we can use the z-score formula:

\[
z = \frac{(X - \mu)}{\sigma}
\]

where:
- \(X\) is the value we are interested in (14 in this case),
- \(\mu\) is the mean (18), and
- \(\sigma\) is the standard deviation (2.7).

Plugging in the values:

\[
z = \frac{(14 - 18)}{2.7} = \frac{-4}{2.7} \approx -1.481
\]

Next, we look up this z-score in the standard normal distribution table (or use a calculator or software) to find the probability of a z-score being less than -1.481.

From the z-table, we find that:

\[
P(Z < -1.481) \approx 0.0694
\]

This represents the proportion of data that is less than 14. To find the proportion of data that is greater than 14, we subtract this value from 1:

\[
P(Z > -1.481) = 1 - P(Z < -1.481) \approx 1 - 0.0694 = 0.9306
\]

Multiplying by 100 to convert to a percentage:

\[
0.9306 \times 100 \approx 93.06\%
\]

So, approximately 93.1% of the data is greater than 14.

Thus, the answer is:

**D. 93.1%**