ms. barnsley separates her class into two groups. she gives each student the same 25 - question math quiz…

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.\nquiz completion times (minutes)\n|group a|4.5|4.6|5.0|4.8|4.4|4.7|5.2|4.6|4.8|4.9|\n|group b|5.5|4.0|4.2|4.8|4.1|3.5|3.9|4.3|4.4|4.1|\nwhich statement is true about the distributions of completion times?\nthe students in group a tended to complete the quiz in less time.\nthe median of group a is greater than the median of group b.\nthe means of both groups are about the same.\nthe standard deviation of group b is less than the standard deviation of group a.

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.\nquiz completion times (minutes)\n|group a|4.5|4.6|5.0|4.8|4.4|4.7|5.2|4.6|4.8|4.9|\n|group b|5.5|4.0|4.2|4.8|4.1|3.5|3.9|4.3|4.4|4.1|\nwhich statement is true about the distributions of completion times?\nthe students in group a tended to complete the quiz in less time.\nthe median of group a is greater than the median of group b.\nthe means of both groups are about the same.\nthe standard deviation of group b is less than the standard deviation of group a.

Answer

Answer:

The students in Group A tended to complete the quiz in less time.

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

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

Step4: Arrange Group B data in ascending order

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

Step5: Calculate median of Group B

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

Step6: Calculate mean of Group B

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

Step7: Analyze statements

  • Mean of Group A ($4.75$) is greater than mean of Group B ($4.28$), so students in Group B tended to complete quiz in less time. This statement is false.
  • Median of Group A ($4.75$) is greater than median of Group B ($4.15$). This statement is true.
  • Mean of Group A is $4.75$ and mean of Group B is $4.28$, they are not about the same. This statement is false.
  • By observing the spread of data, Group B has more spread - out data values (from $3.5$ to $5.5$) compared to Group A (from $4.4$ to $5.2$), so standard deviation of Group B is greater than that of Group A. This statement is false.