Answer to Question #285302 in Statistics and Probability for Zim

Question

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

Answer 1

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

1.

$P_{36}$ class:

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

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

It lies in the class $88-90$

$P_{36}$ class: $87.5-90.5$

The lower boundary point of $87.5-90.5$  is $87.5.$

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

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

$$=87.6385$$

2.

$D_5$ class:

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

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

It lies in the class $88-90$

$D_5$ class: $87.5-90.5$

The lower boundary point of $87.5-90.5$  is $87.5.$

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

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

$$=89.3077$$

3.

$Q_3$ class:

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

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

It lies in the class $91-93$

$Q_3$ class: $90.5-93.5$

The lower boundary point of $90.5-93.5$  is $90.5.$

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

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

$$=92.4375$$