Question

 The time between arrival of 60 patients at an intensive care unit were recorded to the nearest hour. The data are shown below.

Time (hours) 0–19 20–39 40–59 60–79 80–99 100–119 120–139 140–159 160–179

 Frequency     16       13        17       4           4            3              1                1             1

 

Determine the median, mode and variance of the arrivals.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} Time(hours) & Cumulative\ frequency \\ \hline 0 & 0 \\ 19.5 & 16\\ 39.5 & 29\\ 59.5 & 46\\ 79.5 & 50\\ 99.5 & 54\\ 119.5 & 57\\ 139.5 & 58\\ 159.5 & 59\\ 179.5 & 60\\ \end{array}$$

$$mean=\bar{x}=\dfrac{1}{n}\sum_iM_i\cdot f_i$$ $$=\dfrac{1}{60}(9.5(16)+29.5(13)+49.5(17)+69.5(4)$$

$$+89.5(4)+109.5(3)+129.5(1)+149.5(1)$$

$$+169.5(1))=46.5$$

To find Median Class= value of (n/2)th observation=value of (60/2)th observation

=value of (30)th observation

The 30th observation lies in the class $40-59$

The median class is $39.5-59.5$

$n =$ the total frequency$=60,$

$L_m=$ ​is the lower boundary of the class median$=39.5$

$c_f=$ cumulative frequency of the class preceding the median class $=29$

$f=$frequency of the median class $=17$

$c=$ class length of median class $=20$

median:

$$M=L_m+\frac{n/2-c_f}{f}\cdot c$$

$$=39.5+\frac{60/2-29}{17}\cdot 20=40.67647$$

To find Mode Class

Here, maximum frequency is 17.

The mode class is 39.5-59.5.

$L=$ lower boundary point of mode class $=39.5$

$f_1=$  frequency of the mode class $=17$

$f_0=$  frequency of the preceding class $=13$

$f_2=$ frequency of the succeeding class $=4$

$c=$  class length of mode class $=20$

$$Z=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c$$

$$=39.5+(\dfrac{17-13}{2(17)-13-4})\cdot 20$$

$$=44.20588$$

$x=$ the midpoint of class

$A=\dfrac{80+99}{2}=89.5$

$h=$ class length$=20$

$d=\dfrac{A-x}{h}$

variance:

$$\sigma^2=\frac{\sum_i f_id_i^2-(\sum_i f_id_i)^2/n}{n}\cdot h^2$$

$$=\dfrac{1}{n}(\sum_i f_i(A-x_i)^2-\dfrac{1}{n}(\sum_i f_i(A-x_i))^2)$$

$$=\dfrac{1}{60}(\sum_i f_i(A-x_i)^2$$

$$-\dfrac{1}{60}(\sum_i f_i(A-x_i))^2)$$

$$=\dfrac{1}{60}(16(80)^2+13(60)^2+17(40)^2+4(20)^2$$

$$+4(0)^2++3(-20)^2++1(-40)^2+1(-60)^2$$

$$+1(-80)^2-\dfrac{1}{60}(16(80)+13(60)+17(40)$$

$$+4(20)+4(0)+3(-20)+1(-40)+1(-60)$$

$$+1(-80))^2)$$

$$=\dfrac{1}{60}(190800-110940)=1331$$