Question

The top ten scores on a quiz in two classes are listed below.

Class A: 100,100,100,90,95,85,92,98,92,88
Class B: 100,100,98,95,95,96,95,92,81,88
Which statement is true concerning the data?
A)The mode is the same for both classes
B)The range is the same for both classes
C)The mean is the same for both classes
D)The median is the same for both classes

Answer 1

C. the mean is the same for both classes

Explanation:

Answer 2

The mean is the same for both classes

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 =
`\frac{\text{Sum of all observations}}{\text{Total number of observations}}`

Mean =
`(85+88+90+92+92+95+98+100+100+100)/(10)`

Mean = 94

Median = mid value

n = 10 (even)

So, median =
`\frac{(n)/(2)\text{th term}+((n)/(2)+1)\text{th term}}{2}`

=
`\frac{(10)/(2)\text{th term}+((10)/(2)+1)\text{th term}}{2}`

=
`\frac{5\text{th term}+(6)\text{th term}}{2}`

=
`(92+95)/(2)`

=
`93.5`

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 =
`\frac{\text{Sum of all observations}}{\text{Total number of observations}}`

Mean =
`(81+88+92+95+95+95+96+98+100+100)/(10)`

Mean = 94

Median = mid value

n = 10 (even)

So, median =
`\frac{(n)/(2)\text{th term}+((n)/(2)+1)\text{th term}}{2}`

=
`\frac{(10)/(2)\text{th term}+((10)/(2)+1)\text{th term}}{2}`

=
`\frac{5\text{th term}+(6)\text{th term}}{2}`

=
`(95+95)/(2)`

=
`95`

Median = 95

So, The mean is the same for both classes is same i.e. 94.

Hence Option C is correct.