Question

The test scores of 10 students are listed below. 32, 69, 77, 82, 100, 68, 88, 95, 75, 80 a. Determine the five-number summary and draw a boxplot for the given data above. Minimum ________ Q1 ________ Median ________ Q3 ________ Maximum b. Is there any outlier? Justify your answer. c. Which of the measures of center would be best to represent the data? Justify your answer

Answer 1

solution:

Let's Arrange the scores of 10 students in ascending order

32 , 68 , 69 , 75 , 77 , 80 , 82 , 88 , 95 , 100

No.of observations (n) = 10

Five-Number Summary:

---> Minimum value = 32

---> First quartile (Q1)

First quartile (Q1) = (n+2 / 4) th term [since, n is even ]

= (10+2) / 4 th term

= 12/4 th term

= 3rd term

= 69

First quartile (Q1) = 69

---> Median

Median (Q2) = Average of (n/2)th and (n/2)+1 th term

= Average of 5th and 6th terms

= (77+80)/2

= 78.5

Median (Q2) = 78.5

---> Third quartile (Q3)

Third quartile (Q3) = (3n+2)/4 th term [ since, n is even]

= 8th term

= 88

Thirs quartile (Q3) = 88

---> Maximum value = 100

The Boxplot is

b)  

Outliers: The outliers are the data values which are present outside the interval ( Q1-1.5IQR , Q3+1.5IQR)

Here, IQR = Q3-Q1 = 88 - 69 = 19

The interval is : ( 69-1.5(19) , 88+1.5(19) )

: ( 40.5 , 116.5)

Here, observe that the score 32 is present outside the interval

So,32 is only one outlier present in the given test scores

c)

Here , Mean =

= ( 32 +68 +69 +75 +77 +80 +82 +88+95 +100) / 10

= 766/10

= 76.6

Mean = 76.6

observe that : Mean < Median

Hence the data is left skewed and also the presence of outlier effects the mean , which makes the mean as not suitable for skewed data. where as the outliers doesn't effects the median value.

So,Median is the best measure of centre would be best to represent the data