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

b) 95%

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=(x-\mu)/(\sigma)`

`z=(64.2-76.4)/(6.1)`

`z=(-12.2)/(6.1)`

`z=-2`

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

`z=(88.6-76.4)/(6.1)`

`z=(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.