Question

for #8 - 9: the chart shows the times for the 1 mile run for a group of students. work area: 8. how many runners had a faster time than the mean? a) 2 b) 3 c) 4 d) 5 9. how many runners had a faster time than the median? a) 2 b) 3 c) 4 d) 5

Explanation:

Step1: Calculate the mean

The times are 10.2, 10.4, 8.6, 9.7, 9.5, 7. The sum is $10.2 + 10.4+8.6 + 9.7+9.5 + 7=55.4$. There are $n = 6$ data - points. The mean $\bar{x}=\frac{55.4}{6}\approx9.23$.

Step2: Count runners faster than the mean

The times less than 9.23 are 8.6, 7, 9.5. So there are 3 runners faster than the mean.

Step3: Calculate the median

First, order the data: 7, 8.6, 9.5, 9.7, 10.2, 10.4. Since $n = 6$ (an even number), the median $M=\frac{9.5 + 9.7}{2}=9.6$.

Step4: Count runners faster than the median

The times less than 9.6 are 7, 8.6, 9.5. So there are 3 runners faster than the median.

Answer:

1. b) 3
2. b) 3