Question

ms. barnsley separates her class into two groups. she gives each student the same 25 - question math quiz. group a uses a calculator, while group b does not. the table shows the completion times, in minutes, of students in each group. quiz completion times (minutes) group a 4.5 4.6 5.0 4.8 4.4 4.7 5.2 4.6 4.8 4.9 group b 5.5 4.0 4.2 4.8 4.1 3.5 3.9 4.3 4.4 4.1 which statement is true about the distributions of completion times? the students in group a tended to complete the quiz in less time. the median of group a is greater than the median of group b. the means of both groups are about the same. the standard deviation of group b is less than the standard deviation of group a.

Explanation:

Step1: Arrange Group A data in ascending order

$4.4,4.5,4.6,4.6,4.7,4.8,4.8,4.9,5.0,5.2$

Step2: Calculate median of Group A

Since $n = 10$ (even), median$=\frac{4.7 + 4.8}{2}=4.75$

Step3: Calculate mean of Group A

$\bar{x}_A=\frac{4.5 + 4.6+5.0 + 4.8+4.4+4.7+5.2+4.6+4.8+4.9}{10}=\frac{47.5}{10}=4.75$

Step4: Calculate standard - deviation of Group A

First, find the squared - differences from the mean. Then the variance $s_A^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x}_A)^2}{n - 1}$. After calculation, $s_A\approx0.27$

Step5: Arrange Group B data in ascending order

$3.5,3.9,4.0,4.1,4.1,4.2,4.3,4.4,4.8,5.5$

Step6: Calculate median of Group B

Since $n = 10$ (even), median$=\frac{4.1+4.2}{2}=4.15$

Step7: Calculate mean of Group B

$\bar{x}_B=\frac{5.5 + 4.0+4.2+4.8+4.1+3.5+3.9+4.3+4.4+4.1}{10}=\frac{42.8}{10}=4.28$

Step8: Calculate standard - deviation of Group B

First, find the squared - differences from the mean. Then the variance $s_B^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x}_B)^2}{n - 1}$. After calculation, $s_B\approx0.57$

Answer:

The median of Group A is greater than the median of Group B.