Answer to Question #265036 in Statistics and Probability for Vashu

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

median:

$m=L_m+(\frac{n/2-F}{f_m})i$

where n is the total frequency,

F is the cumulative frequency before class median,

f_m is the frequency of the class median,

i is the class width,

L_m is the lower boundary of the class median.

we have:

class median is 40-59

$n=60$

$F=16+13=29$

$f_m=17$

$i=19$

$L_m=40$

$m=40+19\cdot\frac{30-29}{17}=41.12$

mode:

$M=L_{mo}+i\cdot \frac{\Delta_1}{\Delta_1+\Delta_2}$

where L_mo is the lower boundary of class mode,

i is the class width,

$\Delta_1$ is the difference between the frequency of class mode and the frequency of the class after the class mode,

$\Delta_2$ is the difference between the frequency of class mode and the frequency of the class before the class mode.

we have:

class mode is 40-59

$L_{mo}=40$

$\Delta_1=17-4=13$

$\Delta_2=17-13=4$

$M=40+19\cdot\frac{13}{13+4}=54.53$

variance:

$\sigma^2=\frac{\sum fx^2-(\sum fx)^2/n}{n}$

where x is midpoint of class

$\sigma^2=\frac{16\cdot9.5^2+13\cdot29.5^2+17\cdot49.5^2+4\cdot69.5^2+4\cdot89.5^2+3\cdot109.5^2+129.5^2+149.5^2+169.5^2}{60}-$

$-\frac{(16\cdot9.5+13\cdot29.5+17\cdot49.5+4\cdot69.5+4\cdot89.5+3\cdot109.5+129.5+149.5+169.5)^2}{3600}=$

$=3495.25-2162.25=1333$