Question

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Answer 1

The mean of a Trisha's and Martin's datasets is the average values of the datasets

The mean of Martin's scores

Mean is calculated using:

$\bar x = \frac{\sum x}n$

So, we have:

$\bar x = \frac{98+ 78+ 84+ 75+ 91}5$

$\bar x = 85.2$

Hence, the mean of Martin's test scores is 85.2

The mean absolute deviation of Martin's scores

The mean absolute deviation is calculated as:

$MAD = \frac{\sum |x_1 - \bar x|}{n}$

So, we have:

$MAD = \frac{|(98 - 85.2)+ (78- 85.2)+ (84- 85.2)+ (75- 85.2)+ (91- 85.2)|}5$

$MAD = 0$

Hence, the mean absolute deviation of Martin's scores is 0

The mean of Trisha's distances

Mean is calculated using:

$\bar x = \frac{\sum x}n$

So, we have:

$\bar x = \frac{29+ 30+ 28+ 32+ 28+ 31+ 27}7$

$\bar x = 29.3$

Hence, the mean of Trisha's distances is 29.3

The mean absolute deviation of Trisha's distances

The mean absolute deviation is calculated as:

$MAD = \frac{\sum |x_1 - \bar x|}{n}$

So, we have:

$MAD = \frac{|(29 - 29.3)+ (30- 29.3)+ (28- 29.3)+ (32- 29.3)+ (28- 29.3)+ (31- 29.3)+ (27- 29.3)|}7$

$MAD = 0.01$

Hence, the mean absolute deviation of Trisha's distances is 0.01

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