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Difficulty: HardFrequency Distributions, Tables, and Grouped Data

A quality control manager at a pharmaceutical manufacturing facility evaluated the disintegration time, tt (in seconds), for a batch of 400400 coated tablets. The results are summarized in the frequency distribution table below:

Disintegration Time tt (seconds)Frequency
10≤t<2010 \le t < 204040
20≤t<3020 \le t < 30110110
30≤t<4030 \le t < 40160160
40≤t<5040 \le t < 506060
50≤t<6050 \le t < 603030

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

  1. The estimated mean disintegration time of the batch, calculated using class midpoints, is 33.2533.25 seconds.Answer
  2. The median disintegration time of the batch lies within the interval 30≤t<4030 \le t < 40.Answer
  3. C
    Exactly 40%40\% of the tablets in the batch have a disintegration time of less than 3030 seconds.
  4. The ratio of the number of tablets with a disintegration time of at least 3030 seconds to the number of tablets with a disintegration time of less than 3030 seconds is 55 to 33.Answer
  5. E
    More than 25%25\% of the tablets in the batch have a disintegration time of 5050 seconds or greater.

Answer

The statements asserting that the estimated mean is 33.2533.25 seconds, that the median lies in the interval 30≤t<4030 \le t < 40, and that the ratio of tablets taking at least 3030 seconds to those taking less than 3030 seconds is 55 to 33 are all correct.
The estimated mean of 33.2533.25 seconds is correctly computed from class midpoints (13,300/400)(13,300 / 400). The median interval 30≤t<4030 \le t < 40 correctly encompasses the middle values (200th200^{\text{th}} and 201st201^{\text{st}} observations out of 400400). The part-to-part ratio of tablets with disintegration time ≥30\ge 30 seconds (250250) to <30< 30 seconds (150150) simplifies to 250:150=5:3250:150 = 5:3.

Step-by-Step Solution

1
Calculate the estimated mean using class midpoints.
Midpoints are 15,25,35,45,5515, 25, 35, 45, 55. Total product sum ∑f⋅m=(40×15)+(110×25)+(160×35)+(60×45)+(30×55)=13,300\sum f \cdot m = (40 \times 15) + (110 \times 25) + (160 \times 35) + (60 \times 45) + (30 \times 55) = 13,300. Mean =13,300/400=33.25= 13,300 / 400 = 33.25 seconds.
The mean of grouped data is estimated by taking the weighted sum of interval midpoints divided by total sample size.
2
Determine the interval containing the sample median.
Cumulative frequencies: 10≤t<20→4010 \le t < 20 \rightarrow 40; 20≤t<30→15020 \le t < 30 \rightarrow 150; 30≤t<40→31030 \le t < 40 \rightarrow 310. The 200th200^{\text{th}} and 201st201^{\text{st}} data points fall between cumulative counts 150150 and 310310, putting the median in 30≤t<4030 \le t < 40.
The median corresponds to the middle position (N/2=200N/2 = 200) of ordered data.
3
Evaluate the percentage of tablets with disintegration time less than 3030 seconds.
Count =40+110=150= 40 + 110 = 150. Percentage =(150/400)×100%=37.5%= (150 / 400) \times 100\% = 37.5\%.
Part-to-whole percentage requires dividing the sum of frequencies below 3030 seconds by the total sample size 400400.
4
Compute the part-to-part ratio of tablets with t≥30t \ge 30 seconds versus t<30t < 30 seconds.
Tablets with t≥30t \ge 30: 160+60+30=250160 + 60 + 30 = 250. Tablets with t<30t < 30: 40+110=15040 + 110 = 150. Ratio =250:150=5:3= 250 : 150 = 5 : 3.
Formulating the part-to-part ratio comparing the upper three intervals to the lower two intervals.
5
Evaluate the percentage of tablets with disintegration time t≥50t \ge 50 seconds.
Count =30= 30. Percentage =(30/400)×100%=7.5%= (30 / 400) \times 100\% = 7.5\%.
Checking if 7.5%7.5\% exceeds 25%25\% shows the statement is false.

Key Concept

Grouped Data Analysis: Mean Estimation, Median Interval Identification, and Relative Frequency Calculations
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Topics

Frequency Distributions, Tables, and Grouped Data

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