Question

2. Samples of the age in years of student cars and faculty/staff cars: Student: 10,4, 5,2, 9, 7, 8, 8,7, 13, 12, n-11 Faculty: 5, 10, 4, 13, 2,3, 2, 7, 6, 6, 3, 4, n 12 Calculate the mean, median, and mode for each group. Do either of these distributions appear to be skewed? 3. The data below reflect movie viewers' recall of products/brands shown within either humorous or dramatic movies. Calculate the range, variance, and standard deviation of each set of scores Comedies Dramas 4 4

Answer 1

2)

Students :

10,4,5,2,9,7,8,8,7,13,12

Mean = $\frac{\sum x}{n}$ ; $\sum x$ = 10+4+5+2+9+7+8+8+7+13+12 = 85 and n = 11

Mean = 85/11 = 7.7272

Median : It is the middle most number in the "Ordered" data set , that divides the data set into two equal parts.

So first we need to arrange the data set in ascending order.

Since n =11 the middle most number is at 6^th position in the order.

8 is the 6th number in ascending order.

Therefore median = 8

Mode : it is the most frequent or repeated number in the data set.

In the given data 7 and 8 repeated two times ,and other numbers are not repeated

Therefore data does not have any Mode value.

So for student data set has Mean = 7.7272 and Median = 8

Since Mean and Median are approximately same , the distribution of students cars age is symmetric.

Faculty :

Mean = $\frac{\sum x}{n}$ ; $\sum x$ = 5+10+4+13+2+3+2+7+6+6+3+4 = 65 and n = 12

Mean = 65/12 = 5.42

Median :

First we need to arrange the data set in ascending order.

Since n =12 the middle most number is between at 6^th and 7^th position in the order.

4 is the 6th and 5 is at the 7^th number in ascending order.

So we have to take average of both numbers

Median = ( 4+5)/2 = 4.5

Mode :

In the given data 2,3,4 and 6 repeated two times ,and other numbers are not repeated

Therefore data does not have any Mode value.

So for Staff data set has Mean = 5.42 and Median = 4.5

Since Mean > Median , the distribution of faculty cars age is right skewed.

#3)

Comedies :

Range = Maximum number - Minimum number = 6 - 1 = 5

Mean = Mean = $\frac{\sum x}{n}$ ; $\sum x$ =2+3+4+1+6+5= 21 and n = 6

Mean = 21/6 = 3.5

Variance = $\frac{\sum (x-mean)^2}{n-1}$

=$\frac{ (2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(1-3.5)^2+(6-3.5)^2+(5-3.5)^2}{6-1}$

Variance = 3.5

Standard deviation = $\sqrt{Variance}$ = $\sqrt{3.5}$

Standard deviation = 1.87

Dramas :

Range = Maximum number - Minimum number = 6 - 1 = 5

Mean = Mean = $\frac{\sum x}{n}$ ; $\sum x$ =4+2+6+5+8+9 = 34 and n = 6

Mean = 34/6 = 5.67

Variance = $\frac{\sum (x-mean)^2}{n-1}$

=$\frac{ (4-5.67)^2+(2-5.67)^2+(6-5.67)^2+(5-5.67)^2+(8-5.67)^2+(9-5.67)^2}{6-1}$

Variance = 6.67

Standard deviation = $\sqrt{Variance}$ = $\sqrt{6.67}$

Standard deviation = 2.58