Answer to Question #263585 in Statistics and Probability for DANCAN

Question #263585

a)     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.



1
Expert's answer
2021-11-16T18:44:16-0500

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,

fm is the frequency of the class median,

i is the class width,

Lm 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 Lmo 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"

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