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Distributions, Standard Deviations and Z-Scores

4. James Johnson, manager of quality control at Creative Auto Corp., just received a report from the assembly plant. The latest shipment of 200 lug bolts used to attach the wheels showed a mean diameter of 18.01 mm and a median of 17.92 mm. Therefore, James can conclude that the distribution of the diameters of the lug bolts
a. is perfectly symmetric.
b. is skewed left.
c. is skewed right.
d. has a range of 0.09 mm.

5. Juan Salvador just completed a study of the life expectancy of 100 light bulbs and discovered that the mean time that they lasted before burning out was 1,900 hours. If the standard deviation was 150 hours, the empirical rule allows Juan to conclude that approximately 68 of the bulbs burned out between _________ and _________ hours.
a. 1,750, 2,050
b. 1,800, 1,900.
c. 1,600, 2,200.
d. 1,450, 2,350.

Use the following information to answer questions 7 - 8:
Approximately 25% of the population belongs to a health maintenance organization (HMO). Assume that for a randomly selected group of 20 adults, the number belonging to an HMO has a binomial distribution.

7. The probability of finding exactly 5 in the 20 who belong to an HMO is:
a. 0.2023
b. 0.1567
c. 0.1750
d. 0.2345

8. The probability of finding at least one in the 20 who belongs to an HMO is:
a. 0.7500
b. 0.8012
c. 0.9056
d. 0.9968

9. The standard normal table shows an area value of 0.1 for a z-score of 0.25 and an area value of 0.35 for a z-score of 1.04. What percentage of the observations of a random variable that is normally distributed will fall between 0.25 standard deviations below the mean and 1.04 standard deviations above the mean?
a. 25%
b. 35%
c. 45%
d. 55%

10. Trudy Jones recently completed her certification examination and learned that her z-score was -2.5. The examining board also informed her that a failure to pass would be all scores that were 1 or more standard deviations below the mean and that those with scores higher than 2 standard deviations above the mean would receive a special commendation award. Trudy can, therefore, conclude that she
a. failed the exam.
b. passed the exam.
c. passed the exam and will receive a special commendation award.
d. passed the exam, but no commendation award is forthcoming.

Solution Summary

Distributions, Standard Deviations and Z-Scores are investigated. The solution is detailed and well presented.