Question

Difficulty: MediumFrequency Distributions, Tables, and Grouped Data

A environmental monitoring group collected 200200 soil samples from a nature reserve and recorded their pH levels in the frequency table below:

pH Level RangeNumber of Samples
5.05.95.0 – 5.94040
6.06.96.0 – 6.97070
7.07.97.0 – 7.95050
8.08.98.0 – 8.93030
9.09.99.0 – 9.91010

What percent of the soil samples with a pH level of at least 6.06.0 have a pH level in the range 6.06.0 to 7.97.9?

  1. A
    43.75%43.75\%
  2. B
    60.00%60.00\%
  3. 75.00%75.00\%Answer
  4. D
    80.00%80.00\%
  5. E
    68.75%68.75\%

Answer

75.00%75.00\%
To find the desired percentage, first restrict the sample space to all soil samples having a pH level of at least 6.06.0. Adding the frequencies for 6.06.96.0–6.9 (7070), 7.07.97.0–7.9 (5050), 8.08.98.0–8.9 (3030), and 9.09.99.0–9.9 (1010) gives a base total of 160160 samples. Next, find the number of samples within that group that fall in the range 6.06.0 to 7.97.9, which is 70+50=12070 + 50 = 120. The percentage is calculated as (120/160)×100%=75.00%(120 / 160) \times 100\% = 75.00\%.

Step-by-Step Solution

1
Determine the conditional base population (samples with pH 6.0\ge 6.0).
Sum of frequencies for ranges 6.06.96.0-6.9, 7.07.97.0-7.9, 8.08.98.0-8.9, and 9.09.99.0-9.9: 70+50+30+10=16070 + 50 + 30 + 10 = 160.
The question asks 'of the soil samples with a pH level of at least 6.0', which restricts the total base to these four intervals.
2
Determine the target frequency (samples with pH in range 6.06.0 to 7.97.9).
Sum of frequencies for ranges 6.06.96.0-6.9 and 7.07.97.0-7.9: 70+50=12070 + 50 = 120.
This captures all qualifying samples within the specified target interval.
3
Calculate the conditional percentage.
120160×100%=0.75×100%=75.00%\frac{120}{160} \times 100\% = 0.75 \times 100\% = 75.00\%.
Divide the target sub-group count by the conditional base count and convert to a percentage.

Key Concept

Conditional Relative Frequency in Grouped Data Tables
Estimated Time:1m 30s
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