Question

the table shows the maze - completion times, in seconds, of two sets of mice in an experiment.
maze - completion times (seconds)

group a	52	41	45	58	49	55	44	54	48	50
group b	45	62	40	43	57	52	48	60	41	55

which statement is true about the variability of the group times?

• the times in group a have a greater standard deviation.
• the times in group a have a greater median.
• the times in group b have a greater interquartile range.
• the times in group b have a greater mean.

Explanation:

Step1: Calculate Group A mean

$\bar{x}_A = \frac{52+41+45+58+49+55+44+54+48+50}{10} = \frac{496}{10} = 49.6$

Step2: Calculate Group B mean

$\bar{x}_B = \frac{45+62+40+43+57+52+48+60+41+55}{10} = \frac{503}{10} = 50.3$

Step3: Find Group A median

Sort Group A: $41, 44, 45, 48, 49, 50, 52, 54, 55, 58$
Median: $\frac{49+50}{2} = 49.5$

Step4: Find Group B median

Sort Group B: $40, 41, 43, 45, 48, 52, 55, 57, 60, 62$
Median: $\frac{48+52}{2} = 50$

Step5: Calculate Group A IQR

Q1 (Group A): $\frac{44+45}{2}=44.5$, Q3 (Group A): $\frac{54+55}{2}=54.5$
IQR_A = $54.5 - 44.5 = 10$

Step6: Calculate Group B IQR

Q1 (Group B): $\frac{41+43}{2}=42$, Q3 (Group B): $\frac{57+60}{2}=58.5$
IQR_B = $58.5 - 42 = 16.5$

Step7: Evaluate standard deviation (approx)

Group A data clusters closer to mean; Group B has more extreme values (60,62,40), so Group B has larger standard deviation.

Step8: Verify all options

• Option1: False (Group B has larger SD)
• Option2: False (Group B has larger median)
• Option3: True (IQR_B > IQR_A)
• Option4: False ($50.3 > 49.6$, but question asks about variability; this is central tendency, and even so, we check the variability-focused correct option)

Answer:

The times in Group B have a greater interquartile range.