Answer:
The median of Group A is greater than the median of Group B.
Step-by-step explanation:
Given
[tex]\begin{array}{cccccccccc}{Group\ A} & 4.5 & 4.8 & 4.6 & 5.0 & 4.8 & 4.4 & 4.7 & 5.2 & 3.9 \ \\ {Group\ B} & 5.5 & 4.9 & 4.0 & 4.2 & 4.8 & 4.1 & 3.5 & 4.6 & 4.3 \ \end{array}[/tex]
Required
Which of the options is true
(a) Group A complete in less time
To do this, we calculate the average of both using:
[tex]\bar x = \frac{\sum x}{n}[/tex]
Where
[tex]n = 9[/tex]
So, we have:
[tex]\bar x_A = \frac{4.5 + 4.8 + 4.6 + 5.0 + 4.8 + 4.4 + 4.7 + 5.2 + 3.9}{9}[/tex]
[tex]\bar x_A = \frac{41.9}{9}[/tex]
[tex]\bar x_A = 4.66[/tex]
[tex]\bar x_B =\frac{5.5 + 4.9 + 4.0 + 4.2 + 4.8 + 4.1 + 3.5 + 4.6 + 4.3}{9}[/tex]
[tex]\bar x_B =\frac{39.9}{9}[/tex]
[tex]\bar x_B =4.43[/tex]
The average time of Group A is higher than that of B, this means that Group A spend more time, on average.
(b) Group A has a greater median
First, we sort the given data in ascending order
[tex]\begin{array}{cccccccccc}{Group\ A} & 3.9 & 4.4 & 4.5 & 4.6 & 4.7 & 4.8 & 4.8 & 5.0 & 5.2 \ \\ {Group\ B} & 3.5 & 4.0 & 4.1 & 4.2 & 4.3 & 4.6 & 4.8 & 4.9 & 5.5 \ \end{array}[/tex]
The median is then calculated using:
[tex]Median = \frac{n + 1}{2}th[/tex]
This gives:
[tex]Median = \frac{9 + 1}{2}th[/tex]
[tex]Median = \frac{10}{2}th[/tex]
[tex]Median = 5th[/tex]
The median is the fifth item for both groups.
So, we have:
[tex]A =4.7[/tex]
[tex]B =4.3[/tex]
4.7 is greater than 4.3
Hence, (b) is true
Since (b) is true and only one option is correct, then (c) and (d) are incorrect
