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Statistics: Box And Whiskers

Statistics: Box And Whiskers

Assessment

Presentation

Mathematics, Other

11th Grade

Hard

Created by

KASSIA! LLTTF

Used 4+ times

FREE Resource

10 Slides • 0 Questions

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Statistics : Box And Whiskers

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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 : (n+12)thMedian\ OR\ Q2\ :\ \left(\frac{n+1}{2}\right)^{th}  
 Q1=(n+12)th (from 1st set)Q1=\left(\frac{n+1}{2}\right)^{th\ }\left(from\ 1st\ set\right)  
 Q3 =(n+12)th (from 2nd set)Q3\ =\left(\frac{n+1}{2}\right)^{th}\ \left(from\ 2nd\ set\right)  

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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 = (n+12)th = (10+12)th=112th=5.5th Q2\ =\ \left(\frac{n+1}{2}\right)^{th}\ =\ \left(\frac{10+1}{2}\right)^{th}=\frac{11}{2}^{th}=5.5th\    =5th + 6th 2=9+112=202=10=\frac{5th\ +\ 6th\ }{2}=\frac{9+11}{2}=\frac{20}{2}=10  

1st Set - 1, 3 , 5 , 7, 9 ;  n = 5 :
 Q1 = (n+12)th=(5+12)th=62=3rd =5Q1\ =\ \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=(n+12)th=(5+12)th=62=3rd=15Q3=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{5+1}{2}\right)^{th}=\frac{6}{2}=3rd=15  

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ex 2

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

n = 30
 Q2=(n+12)th=(30+12)th=312th=15.5th Q2=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{30+1}{2}\right)^{th}=\frac{31}{2}^{th}=15.5th\    =15th +16th 2=35+352=35=\frac{15th\ +16th\ }{2}=\frac{35+35}{2}=35  


1st set - n = 15 
 Q1=(n+12)th=15+12=162=8th =22Q1=\left(\frac{n+1}{2}\right)^{th}=\frac{15+1}{2}=\frac{16}{2}=8th\ =22  

2nd set - n = 15
 Q3=(n+12)th=15+12=162=8th=39Q3=\left(\frac{n+1}{2}\right)^{th}=\frac{15+1}{2}=\frac{16}{2}=8th=39  

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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.

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Ans

(i) n = 51

 Q2=(n+12)th=51+12=522=26th =71Q2=\left(\frac{n+1}{2}\right)^{th}=\frac{51+1}{2}=\frac{52}{2}=26th\ =71  1st set - n = 25
 Q1=(n+12)th=(25+12)th=262=13th =56Q1=\left(\frac{n+1}{2}\right)^{th}=\left(\frac{25+1}{2}\right)^{th}=\frac{26}{2}=13th\ =56  
2nd set - n = 25 
 Q3=(n+12)th=25+12=262=79Q3=\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. 

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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.

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Ans

(ii) It uses all the values in the data.
(ii) mode =22
(iv) median  =12th +13th 2=30+302=30=\frac{12th\ +13th\ }{2}=\frac{30+30}{2}=30  
(v)  n = 25

 Q2 = 30

1st Set - n =12
 Q3=(n+12)th=12+12=132=7.5thQ3=\left(\frac{n+1}{2}\right)^{th}=\frac{12+1}{2}=\frac{13}{2}=7.5th   =7th+8th2=54+592=56.5=\frac{7th+8th}{2}=\frac{54+59}{2}=56.5  

  Q1=(n+12)th=12+12=7.5thQ1=\left(\frac{n+1}{2}\right)^{th}=\frac{12+1}{2}=7.5th  
 =7th+8th 2=22+222=22=\frac{7th+8th\ }{2}=\frac{22+22}{2}=22  
Quartiles are 22, 30, 56.5

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Statistics : Box And Whiskers

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