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Difficulty: Very hardFrequency Distributions, Tables, and Grouped Data

The frequency distribution table below summarizes the calibration offset errors, xx (in microvolts, μV\mu\text{V}), measured for a sample of 200200 precision voltage sensors in a robotics laboratory.

Offset Error Interval (μV\mu\text{V})Frequency
0≤x<100 \le x < 103232
10≤x<2010 \le x < 20f1f_1
20≤x<3020 \le x < 306868
30≤x<4030 \le x < 40f2f_2
40≤x<5040 \le x < 502424

The estimated mean offset error calculated using the midpoints of the five class intervals is equal to 24.8 μV24.8\ \mu\text{V}. If a sensor is selected at random from among those with an offset error of at least 20 μV20\ \mu\text{V}, what is the probability that its offset error is less than 40 μV40\ \mu\text{V}?

  1. 1417\frac{14}{17}Answer
  2. B
    1425\frac{14}{25}
  3. C
    2531\frac{25}{31}
  4. D
    1134\frac{11}{34}
  5. E
    1725\frac{17}{25}

Answer

The correct answer is 1417\frac{14}{17}, which represents the conditional probability that a sensor's offset error is less than 40 μV40\ \mu\text{V} given that it is at least 20 μV20\ \mu\text{V}.
The correct answer is 1417\frac{14}{17}. Solving the system of equations formed by the total sample size (f1+f2=76f_1 + f_2 = 76) and the estimated midpoint mean (3f1+7f2=4043f_1 + 7f_2 = 404) yields f1=32f_1 = 32 and f2=44f_2 = 44. The number of sensors with offset ≥20 μV\ge 20\ \mu\text{V} is 68+44+24=13668 + 44 + 24 = 136. Among these, the number of sensors with offset <40 μV< 40\ \mu\text{V} is 68+44=11268 + 44 = 112. Thus, the conditional probability is 112136=1417\frac{112}{136} = \frac{14}{17}.

Step-by-Step Solution

1
Set up an equation for the total frequency of the sample.
32+f1+68+f2+24=200  ⟹  f1+f2=7632 + f_1 + 68 + f_2 + 24 = 200 \implies f_1 + f_2 = 76
The sum of all class frequencies must equal the given total sample size of 200.
2
Set up an equation for the estimated mean using interval midpoints.
Midpoints are 5,15,25,35,455, 15, 25, 35, 45. Total weighted sum =32(5)+f1(15)+68(25)+f2(35)+24(45)=160+15f1+1700+35f2+1080=2940+15f1+35f2= 32(5) + f_1(15) + 68(25) + f_2(35) + 24(45) = 160 + 15f_1 + 1700 + 35f_2 + 1080 = 2940 + 15f_1 + 35f_2. Mean =2940+15f1+35f2200=24.8  ⟹  15f1+35f2=2020  ⟹  3f1+7f2=404= \frac{2940 + 15f_1 + 35f_2}{200} = 24.8 \implies 15f_1 + 35f_2 = 2020 \implies 3f_1 + 7f_2 = 404.
The estimated mean of grouped data is the sum of products of interval midpoints and frequencies divided by total sample size.
3
Solve the linear system of equations for f1f_1 and f2f_2.
Multiply f1+f2=76f_1 + f_2 = 76 by 33 to get 3f1+3f2=2283f_1 + 3f_2 = 228. Subtracting from 3f1+7f2=4043f_1 + 7f_2 = 404 gives 4f2=176  ⟹  f2=444f_2 = 176 \implies f_2 = 44. Then f1=76−44=32f_1 = 76 - 44 = 32.
Elimination yields exact unknown frequencies for the remaining intervals.
4
Determine the conditional sample space and target frequency.
Condition (offset ≥20 μV\ge 20\ \mu\text{V}): Intervals [20,30),[30,40),[40,50)[20, 30), [30, 40), [40, 50) with total frequency 68+44+24=13668 + 44 + 24 = 136. Target condition (offset <40 μV< 40\ \mu\text{V} within condition): Intervals [20,30)[20, 30) and [30,40)[30, 40) with frequency 68+44=11268 + 44 = 112.
Conditional probability restricts the denominator to sensors meeting the given condition.
5
Calculate the final probability fraction.
Probability =112136=1417= \frac{112}{136} = \frac{14}{17}.
Dividing target count by conditional total count yields the simplified fraction.

Key Concept

Grouped Data Mean Estimation & Conditional Probability from Frequency Distributions
Estimated Time:2m 30s
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Topics

Frequency Distributions, Tables, and Grouped Data

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