Find the mean ,mode and the median lifetime of the 100 tube lights of the following data lifetime 1000-1099,1100-1199,1200-1299,1300-1399,1400-1499,1500-1599 ,1600-1699,1700-1799,1800-1899 frequency 5,9,11,14,18,17,12,9,5
Answers
Given:
Lifetime 1000-1099,1100-1199,1200-1299,1300-1399,1400-1499,1500-1599 ,1600-1699,1700-1799,1800-1899
Frequency 5,9,11,14,18,17,12,9,5
To find:
Find the mean, mode and the median lifetime of the 100 tube lights of the above data.
Solution:
From given, we have
Lifetime Frequency Mid-point(m) f × m cf
1000-1099 5 1049.5 5247.5 5
1100-1199 9 1149.5 10345.5 14
1200-1299 11 1249.5 13744.5 25
1300-1399 14 1349.5 18893 39
1400-1499 18 1449.5 26091 57
1500-1599 17 1549.5 26341.5 74
1600-1699 12 1649.5 19794 86
1700-1799 9 1749.5 15745.5 95
1800-1899 5 1849.5 9247.5 100
∑f = 100 ∑f × m = 145450
Mean = ∑ f × m / ∑f = 145450/100 = 1454.5
Median = cumulative frequency greater than or equal to ∑f/2
∑f/2 = 100/2 = 50
∴ The median class is 1399.5-1499.5.
Now,
∴L=lower boundary point of median class =1399.5
∴n=Total frequency =100
∴cf=Cumulative frequency of the class preceding the median class =39
∴f=Frequency of the median class =18
∴c=class length of median class =100
Median = 1460.6111
Mode
Here, the maximum frequency is 18.
∴ The mode class is 1399.5-1499.5.
∴L=lower boundary point of mode class =1399.5
∴f1= frequency of the mode class =18
∴f0= frequency of the preceding class =14
∴f2= frequency of the succeeding class =17
∴c= class length of mode class =100
Mode = 1479.5
Answer:
From given, we have
Lifetime Frequency Mid-point(m) f × m cf
1000-1099 5 1049.5 5247.5 5
1100-1199 9 1149.5 10345.5 14
1200-1299 11 1249.5 13744.5 25
1300-1399 14 1349.5 18893 39
1400-1499 18 1449.5 26091 57
1500-1599 17 1549.5 26341.5 74
1600-1699 12 1649.5 19794 86
1700-1799 9 1749.5 15745.5 95
1800-1899 5 1849.5 9247.5 100
∑f = 100 ∑f × m = 145450
Mean = ∑ f × m / ∑f = 145450/100 = 1454.5
Median = cumulative frequency greater than or equal to ∑f/2
∑f/2 = 100/2 = 50
∴ The median class is 1399.5-1499.5.
Now,
∴L=lower boundary point of median class =1399.5
∴n=Total frequency =100
∴cf=Cumulative frequency of the class preceding the median class =39
∴f=Frequency of the median class =18
∴c=class length of median class =100
\begin{gathered}M=L+\dfrac{\frac{n}{2}-cf}{f}\times c\\\\M=1399.5+\dfrac{50-39}{18}\times 100\end{gathered}M=L+f2n−cf×cM=1399.5+1850−39×100
Median = 1460.6111
Mode
Here, the maximum frequency is 18.
∴ The mode class is 1399.5-1499.5.
∴L=lower boundary point of mode class =1399.5
∴f1= frequency of the mode class =18
∴f0= frequency of the preceding class =14
∴f2= frequency of the succeeding class =17
∴c= class length of mode class =100
\begin{gathered}Z=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\times c\\\\Z=1399.5+(\dfrac{18-14}{2(18)-14-17})\times 100\end{gathered}Z=L+(2f1−f0−f2f1−f0)×cZ=1399.5+(2(18)−14−1718−14)×100
Mode = 1479.5