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Difficulty: MediumFrequency Distributions, Tables, and Grouped Data

The frequency distribution table below summarizes the processing times, in minutes, for a sample of 100 health insurance claims processed by an administrative team.

Claim Processing Time (minutes)Frequency
10≤t<2010 \le t < 2015
20≤t<3020 \le t < 3025
30≤t<4030 \le t < 4035
40≤t<5040 \le t < 5020
50≤t<6050 \le t < 605

Which of the following statements regarding this frequency distribution must be true? Select all that apply.

  1. The interval 30≤t<4030 \le t < 40 contains the median processing time for the dataset.Answer
  2. Exactly 60 percent of the claims required 30 minutes or more to process.Answer
  3. C
    The lower bound estimate for the overall mean processing time is 31.0 minutes.
  4. The number of claims in the interval 20≤t<3020 \le t < 30 represents 25 percent of all claims.Answer
  5. E
    The ratio of the number of claims completed in under 30 minutes to the number of claims taking 40 minutes or more is 2 to 1.

Answer

The true statements are: the median processing time lies in the interval 30≤t<4030 \le t < 40; exactly 60 percent of the claims required 30 minutes or more to process; and the number of claims in the interval 20≤t<3020 \le t < 30 represents 25 percent of all claims.
The statement regarding the median is correct because the 50th and 51st values lie within the cumulative frequency span of 41 to 75, corresponding to 30≤t<4030 \le t < 40. The statement regarding claims requiring 30 minutes or more is correct because 35+20+5=6035 + 20 + 5 = 60 out of 100 claims is 60 percent. The statement regarding the 20≤t<3020 \le t < 30 interval is correct because 25 out of 100 claims is exactly 25 percent.

Step-by-Step Solution

1
Calculate cumulative frequencies to locate the median class interval.
Cumulative frequencies are 15 (10≤t<2010 \le t < 20), 40 (20≤t<3020 \le t < 30), 75 (30≤t<4030 \le t < 40), 95 (40≤t<5040 \le t < 50), and 100 (50≤t<6050 \le t < 60). For 100 observations, the 50th and 51st values fall in the 30≤t<4030 \le t < 40 interval.
The median of an even number N=100N=100 of sorted values is the average of the 50th and 51st values.
2
Calculate the percentage of claims taking 30 minutes or more.
Sum the frequencies for intervals ≥30\ge 30: 35+20+5=6035 + 20 + 5 = 60. As a percentage of 100 total claims, this is 60100×100%=60%\frac{60}{100} \times 100\% = 60\%.
To find the proportion meeting or exceeding a threshold, sum all class frequencies with lower limits at or above 30 minutes.
3
Evaluate the percentage representation of the 20≤t<3020 \le t < 30 interval.
Frequency is 25 out of 100 total claims, which equals 25100=25%\frac{25}{100} = 25\%.
The relative frequency of a grouped class is its frequency divided by the total frequency.
4
Determine the lower bound for the estimated mean processing time to evaluate incorrect claims.
Minimum sum = 15(10)+25(20)+35(30)+20(40)+5(50)=275015(10) + 25(20) + 35(30) + 20(40) + 5(50) = 2750. Lower bound mean = 2750/100=27.52750 / 100 = 27.5 minutes.
The absolute minimum mean occurs when every observation equals the lower boundary of its respective interval.
5
Calculate the ratio of claims under 30 minutes to claims 40 minutes or more.
Claims <30< 30 min = 15+25=4015 + 25 = 40. Claims ≥40\ge 40 min = 20+5=2520 + 5 = 25. Ratio = 40:25=8:540 : 25 = 8 : 5.
Must include all relevant intervals (including 50≤t<6050 \le t < 60) when computing the denominator for ≥40\ge 40 minutes.

Key Concept

Grouped Frequency Distributions, Median Class Identification, and Relative Frequencies
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Topics

Frequency Distributions, Tables, and Grouped Data

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