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Difficulty: Very hardNormal Distributions, Standard Deviation Curves, and Percentile Ranks

In a quality control assessment, the weights of manufactured steel components are normally distributed with a mean of 450450 grams and a standard deviation of 1212 grams. Components weighing less than 426426 grams or more than 474474 grams are classified as defective and discarded. Of the remaining non-defective components, those weighing at least 462462 grams are classified as Premium Grade. Assuming the 68–95–99.768\text{--}95\text{--}99.7 empirical rule for normal distributions, approximately how many components in a batch of 10,00010,000 are Premium Grade?

  1. A
    1,2831,283
  2. 1,3501,350Answer
  3. C
    1,5201,520
  4. D
    1,6001,600
  5. E
    3,4003,400

Answer

The correct number of Premium Grade components is 1,3501,350.
By standardizing the given weight thresholds into z-scores (z=−2.0z = -2.0 for 426 g426\text{ g}, z=+1.0z = +1.0 for 462 g462\text{ g}, and z=+2.0z = +2.0 for 474 g474\text{ g}), Premium Grade components are defined by the interval +1.0≤z≤+2.0+1.0 \le z \le +2.0. According to the empirical rule, 95%95\% of data falls within [−2σ,+2σ][-2\sigma, +2\sigma] and 68%68\% falls within [−1σ,+1σ][-1\sigma, +1\sigma]. The portion in the positive tail between +1σ+1\sigma and +2σ+2\sigma is 95%−68%2=13.5%\frac{95\% - 68\%}{2} = 13.5\%. Multiplying 13.5%13.5\% by the batch total of 10,00010,000 yields 1,3501,350 components.

Step-by-Step Solution

1
Calculate the z-scores for the defect thresholds and the Premium Grade threshold.
Lower defect limit: z=426−45012=−2.0z = \frac{426 - 450}{12} = -2.0; Upper defect limit: z=474−45012=+2.0z = \frac{474 - 450}{12} = +2.0; Premium Grade lower threshold: z=462−45012=+1.0z = \frac{462 - 450}{12} = +1.0.
Standardizing the raw weight values into z-scores allows the application of the empirical rule.
2
Identify the z-score interval representing non-defective Premium Grade components.
The target weight interval is 462≤weight≤474462 \le \text{weight} \le 474 grams, corresponding to +1.0≤z≤+2.0+1.0 \le z \le +2.0.
Components must weigh at least 462462 grams (z≥+1.0z \ge +1.0) to be Premium Grade, but must not exceed 474474 grams (z>+2.0z > +2.0) because those exceeding 474474 grams are defective.
3
Determine the percentage of the total distribution within +1.0≤z≤+2.0+1.0 \le z \le +2.0 using the empirical rule.
The area between z=−1.0z = -1.0 and z=+1.0z = +1.0 is 68%68\%, and between z=−2.0z = -2.0 and z=+2.0z = +2.0 is 95%95\%. The region between z=+1.0z = +1.0 and z=+2.0z = +2.0 is 95%−68%2=13.5%\frac{95\% - 68\%}{2} = 13.5\%.
By symmetry of the normal distribution curve, half of the difference between the 2σ2\sigma and 1σ1\sigma intervals lies in the upper tail.
4
Calculate the expected count of Premium Grade components in a batch of 10,00010,000.
10,000×0.135=1,35010,000 \times 0.135 = 1,350.
Multiplying the population proportion by the batch size gives the expected count.

Key Concept

Calculating areas under a normal curve bounded by standard deviation thresholds (z-scores) using the empirical rule.
Estimated Time:2m 30s
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Topics

Normal Distributions, Standard Deviation Curves, and Percentile Ranks

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