Answer :
Answer:
C. the mean is the same for both classes
Step-by-step explanation:
Answer:
The mean is the same for both classes
Step-by-step explanation:
Class A scores of 10 students: 100,100,100,90,95,85,92,98,92,88
Arrange the data in ascending order
85,88,90,92,92,95,98,100,100,100
Mode = 100 since it occurred more times.
Range = Maximum - Minimum = 100-85 = 15
Mean = [tex]\frac{\text{Sum of all observations}}{\text{Total number of observations}}[/tex]
Mean = [tex]\frac{85+88+90+92+92+95+98+100+100+100}{10}[/tex]
Mean = 94
Median = mid value
n = 10 (even)
So, median = [tex]\frac{\frac{n}{2}\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}[/tex]
= [tex]\frac{\frac{10}{2}\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}[/tex]
= [tex]\frac{5\text{th term}+(6)\text{th term}}{2}[/tex]
= [tex]\frac{92+95}{2}[/tex]
= [tex]93.5[/tex]
Median = 93.5
Class B scores of 10 students : 100,100,98,95,95,96,95,92,81,88
Arrange the data in ascending order
81,88,92,95,95,95,96,98,100,100
Mode = 95 since it occurred more times.
Range = Maximum - Minimum = 100-81 = 19
Mean = [tex]\frac{\text{Sum of all observations}}{\text{Total number of observations}}[/tex]
Mean = [tex]\frac{81+88+92+95+95+95+96+98+100+100}{10}[/tex]
Mean = 94
Median = mid value
n = 10 (even)
So, median = [tex]\frac{\frac{n}{2}\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}[/tex]
= [tex]\frac{\frac{10}{2}\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}[/tex]
= [tex]\frac{5\text{th term}+(6)\text{th term}}{2}[/tex]
= [tex]\frac{95+95}{2}[/tex]
= [tex]95[/tex]
Median = 95
So, The mean is the same for both classes is same i.e. 94.
Hence Option C is correct.