Question

Scores on an exam follow an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. Approximately, what percentage of the data is between 64.2 points and 88.6 points?
a) 68%
b) 95%
c) 99.7%
d) 50%

Answer 1

Answer:

b) 95%

Step-by-step explanation:

We have been given that scores on an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. We are asked to find the percentage of the data that is between 64.2 points and 88.6 points.

First of all, we will find z-scores of each data point as:

$z=\frac{x-\mu}{\sigma}$

$z=\frac{64.2-76.4}{6.1}$

$z=\frac{-12.2}{6.1}$

$z=-2$

Let us find z-score corresponding to normal score 88.6.

$z=\frac{88.6-76.4}{6.1}$

$z=\frac{12.2}{6.1}$

$z=2$

To find the percentage of the data is between 64.2 points and 88.6 points, we need to find area under a normal distribution curve that lie within two standard deviation of mean.

The empirical rule of normal distribution states that approximately 95% of data points fall within two standard deviation of mean, therefore, option 'b' is the correct choice.