Answer to Question #247748 in Statistics and Probability for Nora

Question

2. These are the marks of 8 students: 70, 71, 73, 74, 68, 72, 71, 12 a) Calculate the mean. b) Calculate the median. c) Why are they different? 3. What percent of the class scores below each measure of central tendency? % scoring less than mean = % scoring less than the median = 4. Tamara's mark is higher than 63% of the class. a) What percentile does she fall into? b) How many students score less than she does? c) What is Tamara’s score?

Answer 1

2.

$$12,68,70, 71, 71, 72, 73, 74$$

a)

$$n=8, mean=\dfrac{1}{8}(12+68+70+71+71$$

$$+72+73+74)=63.875$$

b)

$$median=\dfrac{71+71}{2}=71$$

c) If the distribution is symmetric then the mean is equal to the median and the distribution will have zero skewness.

Outliers affect the mean value of the data but have little effect on the median or mode.

Since we have outlier 12, the mean and median are different,

3.

$$12<63.875$$

There are 8 scorings. Only one scoring is less than $63.875.$

$$\dfrac{1}{8}\cdot100\%=12.5\%$$

12.5% scoring less than mean.

$$12<71, 68<71, 70<71$$

Three scorings are less than $71.$

$$\dfrac{3}{8}\cdot100\%=37.5\%$$

37.5% scoring less than median.

4.

a) Tamara's mark is higher than 63% of the class. The $k-th$  percentile is the lowest score in the data set that is greater than or equal to a percentage (k) of the scores.

Tamara's percentile rank is 63%.

b)

The lowest five scorings are

$$12,68,70, 71, 71$$

$$\dfrac{5}{8}\cdot100\%=62.5\%<63\%$$

The lowest six scorings are

$$12,68,70, 71, 71, 72$$

$$\dfrac{6}{8}\cdot100\%=75\%>63\%$$

5 students score less than she does.

c)

5 students score less than Tamara does.

Then the sixth scoring in ascending order is Tamara’s score.

Tamara’s score is 72.