Question

Using the same data… 2 3 4 4 4 6 6 6 7 8 8 9 10 10 11 12 16 16 28 46 (d) [5 pts] Determine the 5# summary. (e) Determine the lower and upper fence to determine if there are any outliers. (f) Draw and carefully label a modified boxplot for this data. (g) What is the shape of the distribution (symmetric, skewed left, or skewed right). Explain.

Answer 1

We are given the data set of 20 numbers in the ascending order.

Median is the middle most number , so median is the average of 10th and 11th number in the data set

So Median = (8+8)/2 = 8

Median divides the data set into equal two parts.

1st quartile Q1 is the middle most number of the first half of the data set.

So it is average of 5th and 6th number in the data set.

Q1 = (4+6) / 2

Q1 = 5

3rd quartile Q3 is the middle most number of the 2nd half of the data set.

So it is average of 15th and 16th number in the data set.

Q3 = (11+12) / 2

Q1 = 11.5

d) 5# summary. :  

Minimum = 2

Q1= 5

Median = 8

Q3 = 11.5

Maximum = 46

(e) Determine the lower and upper fence to determine if there are any outliers.

IQR = Q3- Q1 = 11.5 - 5

IQR = 6.5

Lower fence = Q1 - 1.5*IQR = 5 - (1.5*6.5 )

Lower fence = -4.75

Upper fence = Q3 + (1.5*IQR) = 11.5 + (1.5*6.5 )

Upper fence = 21.25

Since the numbers 28 and 46 are greater than 21.25, they are consider as an outliers.

So 28 and 46 are the outlier in the data set.

(f) Draw and carefully label a modified boxplot for this data.

(g) What is the shape of the distribution (symmetric, skewed left, or skewed right). Explain.

Sk =

=

=

Sk = 0.077

Since Sk is greater than 0 , the shape of distribution is skewed right.