Question

one way to measure the duration of subterranean disturbances such as earthquakes and mining is to calculate the root-mean-square time. the following histograms summarize the distributions of the root-mean-square times for two sources of disturbances.

histograms for earthquake disturbances and mining disturbances

based on the histograms, which of the following correctly compares the two distributions?

a the median of the earthquake disturbances is equal to the median of the mining disturbances.

b the median of the earthquake disturbances is less than the median of the mining disturbances.

c the range of the earthquake disturbances is equal to the range of the mining disturbances.

d the range of the earthquake disturbances is less than the range of the mining disturbances

Explanation:

To compare medians: Earthquake's left - skewed (or lower - valued) peak vs Mining's more centered/higher - valued. Earthquake's median is lower. For range: Both have max ~1.6, min ~0 (Earthquake) and ~0.4 (Mining)? Wait, no—wait, Earthquake's min is 0, Mining's min is above 0? Wait, no, the x - axis: Earthquake starts at 0, Mining starts at 0 but first bar at 0.4? Wait, no, the histograms: Earthquake has a bar at 0.0 - 0.2 with high frequency, Mining's first bar is 0.4 - 0.6? Wait, no, looking at the x - axis labels: Earthquake's bins: 0.0, 0.2, 0.4,... Mining's bins: 0.0, 0.2, 0.4,... Wait, no, the first bar for Earthquake is at 0.0 - 0.2 (relative frequency 0.35), for Mining, first bar is 0.4 - 0.6? No, the x - axis ticks: Earthquake: 0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6. Mining: same. But the first bar for Mining: from 0.4? No, the relative frequency for Mining at 0.0 - 0.2 is 0 (since the first bar is at 0.4 - 0.6? Wait, no, the y - axis starts at 0.0. For Earthquake, the first bar (0.0 - 0.2) has rel freq ~0.35. For Mining, the first bar (0.0 - 0.2) has rel freq 0 (since the first bar is at 0.4 - 0.6 with rel freq ~0.2). So Earthquake's data has lower values (starts at 0, more low - end data), Mining's data starts higher. So median: Earthquake's median is in a lower bin than Mining's. So median of earthquake < median of mining. For range: Earthquake's max is ~1.6, min is 0. Mining's max is ~1.6, min is >0 (since first bar at 0.4). So range of earthquake (1.6 - 0 = 1.6) is greater than mining's range (1.6 - min, min >0, so range <1.6). Wait, but the options: Option B says median of earthquake < median of mining. Let's check median calculation. For a histogram, median is where cumulative rel freq is 0.5. Earthquake: first bar (0.0 - 0.2): 0.35. Second bar (0.2 - 0.4): 0.2. Cumulative: 0.35 + 0.2 = 0.55 >0.5. So median in 0.2 - 0.4. Mining: first bar (0.0 - 0.2): 0. Second bar (0.2 - 0.4): 0. Third bar (0.4 - 0.6): 0.2. Fourth bar (0.6 - 0.8): 0.35. Cumulative: 0 + 0 + 0.2+0.35 = 0.55 >0.5. Wait, no, wait Mining's bars: first bar (0.0 - 0.2): 0, second (0.2 - 0.4): 0, third (0.4 - 0.6): ~0.2, fourth (0.6 - 0.8): ~0.35, fifth (0.8 - 1.0): ~0.25, sixth (1.0 - 1.2): ~0.1, seventh (1.2 - 1.4): ~0.1. Wait, cumulative for Mining: 0 (0 - 0.2) + 0 (0.2 - 0.4) + 0.2 (0.4 - 0.6) + 0.35 (0.6 - 0.8) = 0.55 at 0.8. Earthquake: 0.35 (0 - 0.2) + 0.2 (0.2 - 0.4) = 0.55 at 0.4. So Earthquake's median is in 0.2 - 0.4, Mining's in 0.6 - 0.8. So Earthquake's median < Mining's median. So option B is correct.

Answer:

B. The median of the earthquake disturbances is less than the median of the mining disturbances.