Calculate the mean of the following data by all three methods, and compare the results.
Class IntervalFrequency110−1108110−1209120−13011130−14012140−15010 [4 MARKS]
Calculate the mean of the following data by all three methods, and compare the results.
Class IntervalFrequency110−1108110−1209120−13011130−14012140−15010 [4 MARKS]
Concept: 1 Mark
Application: 1 Mark each
Class Intervalxififixi100−1101058840110−12011591035120−130125111375130−140135121620140−150145101450
∑fi=50 ∑fixi=6320
Mean =632050=126.4
Assumed mean method:
Let assumed mean=125
Class intervalxifidifidi100−1101058−20−160110−1201159−10−90120−1301251100130−1401351210120140−1501451020200
∑fi=50∑fidi=70
Mean =125+7050=126.4
Step-deviation method:
Let assumed mean = 125
Class size = 10
Class Intervalxifidiuifiui100−1101058−20−2−16110−1201159−10−1−9120−13012511000130−1401351210112140−1501451020220
∑fi=50 ∑fiui=7
Mean =125+(750×10)=126.4
The mean in all the three methods is equal to 126.4
Concept: 1 Mark
Application: 1 Mark each
Class Intervalxififixi100−1101058840110−12011591035120−130125111375130−140135121620140−150145101450
∑fi=50 ∑fixi=6320
Mean =632050=126.4
Assumed mean method:
Let assumed mean=125
Class intervalxifidifidi100−1101058−20−160110−1201159−10−90120−1301251100130−1401351210120140−1501451020200
∑fi=50∑fidi=70
Mean =125+7050=126.4
Step-deviation method:
Let assumed mean = 125
Class size = 10
Class Intervalxifidiuifiui100−1101058−20−2−16110−1201159−10−1−9120−13012511000130−1401351210112140−1501451020220
∑fi=50 ∑fiui=7
Mean =125+(750×10)=126.4
The mean in all the three methods is equal to 126.4
