5-4. Exercises

1. BASIC

Ex 1. Find the value of z∗z^* that yields the probability shown.

  1. P(Z<z∗)=0.0075P(Z<z^*)=0.0075

  2. P(Z<z∗)=0.9850P(Z<z^*)=0.9850

  3. P(Z>z∗)=0.8997P(Z>z^*)=0.8997

  4. P(Z>z∗)=0.0110P(Z>z^*)=0.0110

Ex 2. Find the value of z∗z^* that yields the probability shown.

  1. P(Z<z∗)=0.3300P(Z<z^*)=0.3300

  2. P(Z<z∗)=0.9901P(Z<z^*)=0.9901

  3. P(Z>z∗)=0.0055P(Z>z^*)=0.0055

  4. P(Z>z∗)=0.7995P(Z>z^*)=0.7995

Ex 3. Find the value of z∗z^* that yields the probability shown.

  1. P(Z<z∗)=0.1500P(Z<z^*)=0.1500

  2. P(Z<z∗)=0.7500P(Z<z^*)=0.7500

  3. P(Z>z∗)=0.3333P(Z>z^*)=0.3333

  4. P(Z>z∗)=0.8000P(Z>z^*)=0.8000

Ex 4. Find the value of z∗z^* that yields the probability shown.

  1. P(Z<z∗)=0.2200P(Z<z^*)=0.2200

  2. P(Z<z∗)=0.6000P(Z<z^*)=0.6000

  3. a=ba = b

  4. P(Z>z∗)=0.8200P(Z>z^*)=0.8200

Ex 5. Find the indicated value of ZZ . (It is easier to find −zc−z_c and negate it.)

  1. z0.025

  2. z0.20

Ex 6. Find the indicated value of ZZ. (It is easier to find −zc−z_c and negate it.)

  1. z0.002

  2. z0.02

Ex 7. Find the value of x∗x^* that yields the probability shown, where XX is a normally distributed random variable XX with mean 83 and standard deviation 4.

  1. P(X<x*)=0.8700P(X<x*)=0.8700

  2. P(X>x*)=0.0500P(X>x*)=0.0500

Ex 8. Find the value of x∗x^* that yields the probability shown, where XX is a normally distributed random variable XX with mean 54 and standard deviation 12.

  1. P(X<x*)=0.0900P(X<x*)=0.0900

  2. P(X>x*)=0.6500P(X>x*)=0.6500

Ex 9. XX is a normally distributed random variable XX with mean 15 and standard deviation 0.25. Find the values xLx_L and xRx_R of XX that are symmetrically located with respect to the mean of XX and satisfy P(xL<X<xR)=0.80P(x_L < X < x_R) = 0.80 . (Hint. First solve the corresponding problem for ZZ.)

Ex 10. XX is a normally distributed random variable XX with mean 28 and standard deviation 3.7. Find the values xLx_L and xRx_R of XX that are symmetrically located with respect to the mean of XXand satisfy P(xL<X<xR)=0.65P(x_L < X < x_R) = 0.65. (Hint. First solve the corresponding problem for ZZ.)

2. APPLICATIONS

Ex 11. Scores on a national exam are normally distributed with mean 382 and standard deviation 26.

  1. Find the score that is the 50th percentile.

  2. Find the score that is the 90th percentile.

Ex 12. Heights of women are normally distributed with mean 63.7 inches and standard deviation 2.47 inches.

  1. Find the height that is the 10th percentile.

  2. Find the height that is the 80th percentile.

Ex 13. The monthly amount of water used per household in a small community is normally distributed with mean 7,069 gallons and standard deviation 58 gallons. Find the three quartiles for the amount of water used.

Ex 14. The quantity of gasoline purchased in a single sale at a chain of filling stations in a certain region is normally distributed with mean 11.6 gallons and standard deviation 2.78 gallons. Find the three quartiles for the quantity of gasoline purchased in a single sale.

Ex 15. Scores on the common final exam given in a large enrollment multiple section course were normally distributed with mean 69.35 and standard deviation 12.93. The department has the rule that in order to receive an AA in the course his score must be in the top 10% of all exam scores. Find the minimum exam score that meets this requirement.

Ex 16. The average finishing time among all high school boys in a particular track event in a certain state is 5 minutes 17 seconds. Times are normally distributed with standard deviation 12 seconds.

  1. The qualifying time in this event for participation in the state meet is to be set so that only the fastest 5% of all runners qualify. Find the qualifying time. (Hint: Convert seconds to minutes.)

  2. In the western region of the state the times of all boys running in this event are normally distributed with standard deviation 12 seconds, but with mean 5 minutes 22 seconds. Find the proportion of boys from this region who qualify to run in this event in the state meet.

Ex 17. Tests of a new tire developed by a tire manufacturer led to an estimated mean tread life of 67,350 miles and standard deviation of 1,120 miles. The manufacturer will advertise the lifetime of the tire (for example, a “50,000 mile tire”) using the largest value for which it is expected that 98% of the tires will last at least that long. Assuming tire life is normally distributed, find that advertised value.

Ex 18. Tests of a new light led to an estimated mean life of 1,321 hours and standard deviation of 106 hours. The manufacturer will advertise the lifetime of the bulb using the largest value for which it is expected that 90% of the bulbs will last at least that long. Assuming bulb life is normally distributed, find that advertised value.

Ex 19. The weights XX of eggs produced at a particular farm are normally distributed with mean 1.72 ounces and standard deviation 0.12 ounce. Eggs whose weights lie in the middle 75% of the distribution of weights of all eggs are classified as “medium.” Find the maximum and minimum weights of such eggs. (These weights are endpoints of an interval that is symmetric about the mean and in which the weights of 75% of the eggs produced at this farm lie.)

Ex 20. The lengths XX of hardwood flooring strips are normally distributed with mean 28.9 inches and standard deviation 6.12 inches. Strips whose lengths lie in the middle 80% of the distribution of lengths of all strips are classified as “average-length strips.” Find the maximum and minimum lengths of such strips. (These lengths are endpoints of an interval that is symmetric about the mean and in which the lengths of 80% of the hardwood strips lie.)

Ex 21. All students in a large enrollment multiple section course take common in-class exams and a common final, and submit common homework assignments. Course grades are assigned based on students' final overall scores, which are approximately normally distributed. The department assigns a CC to students whose scores constitute the middle 2/3 of all scores. If scores this semester had mean 72.5 and standard deviation 6.14, find the interval of scores that will be assigned a CC.

Ex 22. Researchers wish to investigate the overall health of individuals with abnormally high or low levels of glucose in the blood stream. Suppose glucose levels are normally distributed with mean 96 and standard deviation 8.5 mg/ dℓd ℓ , and that “normal” is defined as the middle 90% of the population. Find the interval of normal glucose levels, that is, the interval centered at 96 that contains 90% of all glucose levels in the population.

3. ADDITIONAL EXERCISES

Ex 23. A machine for filling 2-liter bottles of soft drink delivers an amount to each bottle that varies from bottle to bottle according to a normal distribution with standard deviation 0.002 liter and mean whatever amount the machine is set to deliver.

  1. If the machine is set to deliver 2 liters (so the mean amount delivered is 2 liters) what proportion of the bottles will contain at least 2 liters of soft drink?

  2. Find the minimum setting of the mean amount delivered by the machine so that at least 99% of all bottles will contain at least 2 liters.

Ex 24. A nursery has observed that the mean number of days it must darken the environment of a species poinsettia plant daily in order to have it ready for market is 71 days. Suppose the lengths of such periods of darkening are normally distributed with standard deviation 2 days. Find the number of days in advance of the projected delivery dates of the plants to market that the nursery must begin the daily darkening process in order that at least 95% of the plants will be ready on time. (Poinsettias are so long-lived that once ready for market the plant remains salable indefinitely.)

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