Question #285302

The efficacy of rating of 155 faculty members of a certain college was taken and shown below.


Classes f

73-75 2

76-78 6

79-81 12

82-84 16

85-87 18

88-90 39

91-93 36

94-96 21

97-99 5


Determine the value of the following:

1. P36

2. D5

3. Q3




1
Expert's answer
2022-01-07T13:02:14-0500
ClassfiClasscfboundaries7375272.575.527678675.578.5879811278.581.52082841681.584.53685871884.587.55488903987.590.59391933690.593.512994962193.596.51509799596.599.5155n=155\def\arraystretch{1.5} \begin{array}{c:c:c:c} Class & f_i & Class & cf \\ & & boundaries & & & & \\ \hline 73-75 & 2 & 72.5-75.5 & 2 \\ \hdashline 76-78 & 6 & 75.5-78.5 & 8 \\ \hdashline 79-81 & 12 & 78.5-81.5 & 20 \\ \hdashline 82-84 & 16 & 81.5-84.5 & 36 \\ \hdashline 85-87 & 18 & 84.5-87.5 & 54 \\ \hdashline 88-90 & 39 & 87.5-90.5 & 93 \\ \hdashline 91-93 & 36 & 90.5-93.5 & 129 \\ \hdashline 94-96 & 21 & 93.5-96.5 & 150 \\ \hdashline 97-99 & 5 & 96.5-99.5 & 155 \\ \hdashline & n=155 & & & & \\ \end{array}


1.

P36P_{36} class:

Class with (36n100)th(\dfrac{36n }{100})^{th} value of the observation in cfcf  column

(36(155)100)th(\dfrac{36(155) }{100})^{th} value =(55.8)th=(55.8)^{th} value

It lies in the class 889088-90

P36P_{36} class: 87.590.587.5-90.5

The lower boundary point of 87.590.587.5-90.5  is 87.5.87.5.

L=L= ​lower boundary point of median class=87.5=87.5

P36=L+36n100cffc=87.5+55.854393P_{36}=L+\dfrac{\dfrac{36n}{100}-cf}{f}\cdot c=87.5+\dfrac{55.8-54}{39}\cdot 3

=87.6385=87.6385

2.

D5D_5 class:

Class with (5n10)th(\dfrac{5n }{10})^{th} value of the observation in cfcf  column

(5(155)10)th(\dfrac{5(155) }{10})^{th} value =(77.5)th=(77.5)^{th} value

It lies in the class 889088-90

D5D_5 class: 87.590.587.5-90.5

The lower boundary point of 87.590.587.5-90.5  is 87.5.87.5.

L=L= ​lower boundary point of median class=87.5=87.5

D5=L+5n10cffc=87.5+77.554393D_5=L+\dfrac{\dfrac{5n}{10}-cf}{f}\cdot c=87.5+\dfrac{77.5-54}{39}\cdot 3

=89.3077=89.3077



3.

Q3Q_3 class:

Class with (3n4)th(\dfrac{3n }{4})^{th} value of the observation in cfcf  column

(3(155)4)th(\dfrac{3(155) }{4})^{th} value =(166.25)th=(166.25)^{th} value

It lies in the class 919391-93

Q3Q_3 class: 90.593.590.5-93.5

The lower boundary point of 90.593.590.5-93.5  is 90.5.90.5.

L=L= ​lower boundary point of median class=90.5=90.5


Q3=L+3n4cffc=90.5+116.2593363Q_3=L+\dfrac{\dfrac{3n}{4}-cf}{f}\cdot c=90.5+\dfrac{116.25-93}{36}\cdot 3

=92.4375=92.4375

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