Box and Whisker Plot and Stem and Leaf

Statistics : Box And Whiskers

The box and whiskers plot is another way of showing spread. It uses 5 Values only . * Minimum Value * Maximum value * Lower Quartile (Q1) * Upper Quartile (Q3) * Median (Q2). It has a scale (the range for which is based on the min and max values in the raw data)  . The quartiles are calculated. The minimum and maximum values are taken straight from the raw data.
n = no. of data 
Formulas :

IQR = Q3 - Q1
 $Median\ OR\ Q2\ :\ \left(\frac{n+1}{2}\right)^{th}$  
 $Q1=\left(\frac{n+1}{2}\right)^{th\ }\left(from\ 1st\ set\right)$  
 $Q3\ =\left(\frac{n+1}{2}\right)^{th}\ \left(from\ 2nd\ set\right)$  

ex 1

Find the quartiles and draw a box and whisker for the following data.
1, 3, 5, 7, 9, 11, 13, 15, 17, 19

n = 10

 $Q2\ =\ \left(\frac{n+1}{2}\right)^{th}\ =\ \left(\frac{10+1}{2}\right)^{th}=\frac{11}{2}^{th}=5.5th\$   $=\frac{5th\ +\ 6th\ }{2}=\frac{9+11}{2}=\frac{20}{2}=10$  

1st Set - 1, 3 , 5 , 7, 9 ;  n = 5 :
 $Q1\ =\ \left(\frac{n+1}{2}\right)^{th}=\left(\frac{5+1}{2}\right)^{th}=\frac{6}{2}=3rd\ =5$  
2nd Set - 11, 13, 15, 17 , 19 ; n = 5 : 
 $Q3=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{5+1}{2}\right)^{th}=\frac{6}{2}=3rd=15$  

ex 2

Use the stem and leaf stem plot to find the quartiles and hence draw and box and whiskers plot.

n = 30
 $Q2=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{30+1}{2}\right)^{th}=\frac{31}{2}^{th}=15.5th\$   $=\frac{15th\ +16th\ }{2}=\frac{35+35}{2}=35$  

1st set - n = 15 
 $Q1=\left(\frac{n+1}{2}\right)^{th}=\frac{15+1}{2}=\frac{16}{2}=8th\ =22$  

2nd set - n = 15
 $Q3=\left(\frac{n+1}{2}\right)^{th}=\frac{15+1}{2}=\frac{16}{2}=8th=39$  

2014 No. 7(c)

The aptitude scores obtained by 51 applicants for a supervisory job are summarized in the following stem and leaf diagram.

(i) Find the median and quartiles for the data given.

(ii) Construct a box-and-whisker plot to illustrate the data given and comment on the distribution of the data.

Ans

(i) n = 51

 $Q2=\left(\frac{n+1}{2}\right)^{th}=\frac{51+1}{2}=\frac{52}{2}=26th\ =71$  1st set - n = 25
 $Q1=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{25+1}{2}\right)^{th}=\frac{26}{2}=13th\ =56$  
2nd set - n = 25 
 $Q3=\left(\frac{n+1}{2}\right)^{th}=\frac{25+1}{2}=\frac{26}{2}=79$  
(ii) The second set of applicants performed better / scored higher than the first set of applicants. 

2016 7(a)

(a) Use the data set provided below to answer the questions which follow.

(i) Construct a stem-and-leaf diagram to represent the given data.

(ii) State an advantage of using the stem - and - leaf diagram to represent the given data.

(iii) Determine the mode.

(iv) Determine the median.

(v) Determine the interquartile range.

Ans

(ii) It uses all the values in the data.
(ii) mode =22
(iv) median  $=\frac{12th\ +13th\ }{2}=\frac{30+30}{2}=30$  
(v)  n = 25

 Q2 = 30

1st Set - n =12
 $Q3=\left(\frac{n+1}{2}\right)^{th}=\frac{12+1}{2}=\frac{13}{2}=7.5th$   $=\frac{7th+8th}{2}=\frac{54+59}{2}=56.5$  

  $Q1=\left(\frac{n+1}{2}\right)^{th}=\frac{12+1}{2}=7.5th$  
 $=\frac{7th+8th\ }{2}=\frac{22+22}{2}=22$  
Quartiles are 22, 30, 56.5