Question

Ten students took a placement test to see if they could bypass a prerequisite requirement for a course. Out of 100 points possible, the scores were: 45, 62, 63, 58, 81, 77, 64, 69, 82, 51. The student who scored 69 was at what percentile?

Answer 1

Answer:

$z= \frac{69 -65.2}{11.68}=0.325$

And we can find the probability:

$P(Z$

And that correcpond to the 63 percentile approximated.

Step-by-step explanation:

We have the following dataset:

45, 62, 63, 58, 81, 77, 64, 69, 82, 51.

In order to find this problem we need to assume a distribution for the data. If we assume that the scores are normally distributed we can use the z score in order to find the percentile for the 69 value.

First we need to find the mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}=65.2$

$\sigma=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n}}=11.68$

And then we can find the z score with the following formula:

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

If we replace for X =69 we got this:

$z= \frac{69 -65.2}{11.68}=0.325$

And we can find the probability:

$P(Z$

And that correcpond to the 63 percentile approximated.