Answer to Question #130554 in Statistics and Probability for Samuel kassapa

Question

In 1984, the 51 states in a country recorded a mean and median dropout rate of 26.5% and 25.3% respectively. Interpret the 1984 median dropout rate.

If a test was generally very easy, except for a few students who had very low scores, then the shape of the distribution of scores would be .......

Find the standard deviation of the set of data below.
y + 1, y + 2, y + 5, y + 9, y + 8.


Many professional schools require applicants to take a standardized test. Suppose 1000 students write the test, and your mark of 63 (out of 100) was the 73rd percentile. What does this mean?

Answer 1

solution 1

The median (25.3%) is smaller than the mean (26.5%). This indicates the population's drop out rate is skewed to the right.

solution 2

the distribution of scores will be skewed to the left

solution 3

$$Variance, \sigma^2=\frac{\sum (x_i - \bar x)^2}{n}$$

$$\bar x =\frac{ \sum x_i}{n}$$

$$= \frac{y + 1+y + 2+y + 5+ y + 9+y + 8}{5}=y + 5$$

Therefore

$$\sigma^2 = \frac{16+9+0+16+9}{5} =10$$

Standard deviation

$\sigma =\sqrt{10} = 3.1623$

answer: 3.1623

Solution 4

73rd percentile means you performed better than 73% of the students i.e. 730 out of 1000 students who sat the test got less than 63%