Question #272891

The table below shows the marks scored by

229

students in a statistics class test.


Marks

10  20 20  30 30  40 40  50 50  60 60  70 70 80

Frequency

12 30 x 65 y 25 17

 Given that the median mark is

46

, determine the:

(i) Values of

x

and

y

; [5 Marks]

(ii) Mode; [2 Marks]

(iii) Mean absolute deviation; [4 Marks]

(iv) Quartile deviation. [4 Marks]


1
Expert's answer
2021-11-30T16:48:34-0500

i)

for grouped data:

median = L+w(n/2B)GL+\frac{w(n/2-B)}{G}

where:

  • L is the lower class boundary of the group containing the median
  • n is the total number of values
  • B is the cumulative frequency of the groups before the median group
  • G is the frequency of the median group
  • w is the group width


median group: 40-50

cumulative frequency of median class: B=12+30+xB=12+30+x

then:

median = 46=40+10(229/2(12+30+x))6246=40+\frac{10(229/2-(12+30+x))}{62}


x=229/2662/101230=35x=229/2-6\cdot62/10-12-30=35

y=n123035652517=229123035652517=45y=n-12-30-35-65-25-17=229-12-30-35-65-25-17=45


ii)

Mode = L+fmfm1(fmfm1)+(fmfm+1)wL + \frac{f_m − f_{m-1}}{(f_m − f_{m-1}) + (f_m − f_{m+1})} w

where:

  • L is the lower class boundary of the modal group
  • fm-1 is the frequency of the group before the modal group
  • fm is the frequency of the modal group
  • fm+1 is the frequency of the group after the modal group
  • w is the group width


modal group: 40-50


Mode = 40+10(6535)6535+6545=4640 + \frac{10(65 − 35)}{65-35 +65-45} =46


iii)

Mean absolute deviation:

dm=xixfindm=\frac{\sum |x_i-\overline{x}|f_i}{n}

where fi is frequency,

xi is midpoint of class,

mean:

x=fixin=1512+2530+3535+4565+5545+6525+7517229=45.7\overline{x}=\frac{\sum f_ix_i}{n}=\frac{15\cdot12+25\cdot30+35\cdot35+45\cdot65+55\cdot45+65\cdot25+75\cdot17}{229}=45.7


dm=30.712+20.730+10.735+0.765+9.345+19.325+29.317229=12.26dm=\frac{30.7\cdot12+20.7\cdot30+10.7\cdot35+0.7\cdot65+9.3\cdot45+19.3\cdot25+29.3\cdot17}{229}=12.26


iv)

 Quartile deviation:

QD=(Q3Q1)/2QD=(Q_3-Q_1)/2


Qi class = (in/4)th value of the observation


Qi=L+in/4cffwQ_i=L+\frac{in/4-cf}{f}⋅w

where cf is cumulative frequency


Q1 class = (229/4)=(57.25)th value of the observation

class: 30-40


Q1=30+57.25773510=24.36Q_1=30+\frac{57.25-77}{35}⋅10=24.36


Q3 class = (2293/4229\cdot3/4 )=(171.75)th value of the observation

class: 50-60


Q1=50+171.751874510=46.61Q_1=50+\frac{171.75-187}{45}⋅10=46.61


QD=(46.6124.36)/2=11.13QD=(46.61-24.36)/2=11.13



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