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Difficulty: Very hardFrequency Distributions, Tables, and Grouped Data

A reliability study recorded the operating lifespan, tt (in thousands of hours), for a sample of 250250 semiconductor laser diodes. The results are summarized in the grouped frequency table below.

Lifespan tt (thousands of hours)Frequency
0≤t<40 \le t < 43535
4≤t<84 \le t < 85555
8≤t<128 \le t < 128080
12≤t<1612 \le t < 165050
16≤t<2016 \le t < 203030

Laser diodes with an operating lifespan of at least 60006{}000 hours (t≥6t \ge 6) but less than 1400014{}000 hours (t<14t < 14) are designated as high-efficiency units. Assuming that the values within each class interval are uniformly distributed, what percentage of the 250250 laser diodes in the sample are designated as high-efficiency units?

Answer: 53 %

Answer

53%
To find the percentage of diodes with lifespans between 60006{}000 and 1400014{}000 hours (6≤t<146 \le t < 14), evaluate the relevant intervals. For 4≤t<84 \le t < 8, the sub-interval [6,8)[6, 8) represents 8−68−4=0.5\frac{8-6}{8-4} = 0.5 of the interval width, containing 0.5×55=27.50.5 \times 55 = 27.5 diodes. The interval [8,12)[8, 12) is fully contained, contributing 8080 diodes. For 12≤t<1612 \le t < 16, the sub-interval [12,14)[12, 14) represents 14−1216−12=0.5\frac{14-12}{16-12} = 0.5 of the interval width, containing 0.5×50=250.5 \times 50 = 25 diodes. Totaling these gives 27.5+80+25=132.527.5 + 80 + 25 = 132.5 diodes. Expressed as a percentage of the total 250250 diodes, 132.5250×100%=53%\frac{132.5}{250} \times 100\% = 53\%.

Step-by-Step Solution

1
Determine the estimated number of diodes in the partial interval 6≤t<86 \le t < 8.
The target range [6,8)[6, 8) covers half of the interval [4,8)[4, 8) width of 44 units. With uniform distribution, the count is 0.5×55=27.50.5 \times 55 = 27.5 diodes.
Linear interpolation estimates frequencies for sub-intervals within grouped data.
2
Include the count for the complete interval 8≤t<128 \le t < 12.
All 8080 diodes in this interval fall within 6≤t<146 \le t < 14.
The entire interval is fully contained within the target upper and lower bounds.
3
Determine the estimated number of diodes in the partial interval 12≤t<1412 \le t < 14.
The target range [12,14)[12, 14) covers half of the interval [12,16)[12, 16) width of 44 units. With uniform distribution, the count is 0.5×50=250.5 \times 50 = 25 diodes.
Linear interpolation estimates frequencies for the upper partial boundary.
4
Sum the target diode counts and convert to a percentage of the total sample.
Total target count = 27.5+80+25=132.527.5 + 80 + 25 = 132.5 diodes. Percentage = 132.5250×100%=53%\frac{132.5}{250} \times 100\% = 53\%.
Divide the calculated frequency sum by the total sample size of 250 and multiply by 100.

Key Concept

Grouped Frequency Distribution and Linear Interpolation
Estimated Time:2m 0s
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Topics

Frequency Distributions, Tables, and Grouped Data

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