The following data given the information on the observed life terms -in hours- of - 200- electrical components-Life times -in hours- -9-40-6060-8080-100100-120120-140140-160Frequency- 25- 38- 65- 24- 31- 17-Determine the modal lifetimes of the components-

#160;Life time #160;Frequency #160;40 60 #160;25 #160;60 80 #160;38 #160;80 100 #160;65 #160;100 120 #160;24 #160;120 140 #160;31 #16 ...

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Byju's Answer

Standard XI

Statistics

Mode of Continuous Series

The following...

Question

The following data given the information on the observed life terms (in hours) of $200$ electrical components.
Life times (in hours) 40-60 60-80 80-100 100-120 120-140 140-160
Frequency
$25$ $38$ $65$ $24$ $31$ $17$
Determine the modal lifetimes of the components.

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Solution

Life time Frequency
40-60 25
60-80 38
80-100 65
100-120 24
120-140 31
140-160 17
Class size $h = 20$
$l = 60$
$f = 65$
$f_{1} = 38$
$f_{2} = 24$
$M o d e = l + \frac{f_{1} - f_{2}}{2 f - f_{1} - f_{2}} \times h M = 60 + \frac{38 - 24}{2 (65) - 28 - 24} \times 20 M = 60 + 3.58 M = 63.58$

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Life times (in hours)	40-60	60-80	80-100	100-120	120-140	140-160
Frequency	$25$	$38$	$65$	$24$	$31$	$17$

Life time	Frequency
40-60	25
60-80	38
80-100	65
100-120	24
120-140	31
140-160	17

The following data given the information on the observed life terms (in hours) of 200 electrical components.Life times (in hours) 40-6060-8080-100100-120120-140140-160Frequency253865243117Determine the modal lifetimes of the components.

Life time Frequency 40-60 25 60-80 38 80-100 65 100-120 24 120-140 31 140-160 17Class size h=20l=60f=65f1=38f2=24Mode=l+f1−f22f−f1−f2×hM=60+38−242(65)−28−24×20M=60+3.58M=63.58

The following data given the information on the observed life terms (in hours) of $200$ electrical components.
Life times (in hours) 40-60 60-80 80-100 100-120 120-140 140-160
Frequency
$25$ $38$ $65$ $24$ $31$ $17$
Determine the modal lifetimes of the components.

Life time Frequency
40-60 25
60-80 38
80-100 65
100-120 24
120-140 31
140-160 17
Class size $h = 20$
$l = 60$
$f = 65$
$f_{1} = 38$
$f_{2} = 24$
$M o d e = l + \frac{f_{1} - f_{2}}{2 f - f_{1} - f_{2}} \times h M = 60 + \frac{38 - 24}{2 (65) - 28 - 24} \times 20 M = 60 + 3.58 M = 63.58$