CHAPTER 2 REVIEW



CHAPTER 2 REVIEW

1. The angiogram is a standard diagnostic test used in clinical medicine to detect stroke in patients. This test has some risks for the patient, and several noninvasive techniques have been developed that are hoped to be as effective as the angiogram. One such method utilizes that measurement of cerebral blood flow (CBF) in the brain, since stroke patients tend to have lower levels of CBF than normal. Among healthy people, CBF is normally distributed with mean 75 and standard devitation 17. Patients are classified as being at risk for stroke if their CBF is below 40. What proportion of normal patients will be mistakenly classified as being at risk for stroke?

ANS: Find the area left of Z = -2.06. The answer is: 0.0197.

2. Maple tree diameters in a forest area are normally distributed with mean 10 inches and standard deviation 2.2 inches. Find the proportion of trees having a diameter greater than 15 inches.

ANS: Find the area right of Z = 2.27. The answer is: 0.0116.

3. Our subjects are 35-44-year-old males whose blood pressures are normally distributed with mean 80 and standard deviation 12; N(80, 12). A borderline hypertensive is defined as a person whose diastolic blood pressure is between 90 and 95 mm Hg inclusive; what proportion of subjects are borderline hypertensive? A hypertensive is a person whose diastolic blood pressure is above 95 mm Hg; what proportion of subjects are hypertensive?

ANS: Find the area between Z = 0.83 and Z = 1.25. The answer is 0.8944 - 0.7967 = 0.0977.

4. White blood cell (WBC) count per cubic millimeter of whole blood has approximately the N(7500, 1750) distribution. The lowest 2% of all WBC counts are defined to be probable risks. How low must one's WBC count be to fall in the at-risk group?

ANS: The Z value with 2% = 0.02 area left of it is (approximately) -2.05. So, we need the WBC count that is 2.05 times the standard deviation below (because the Z-score is negative) the mean. The answer is 7500 - 2.05 * 1750 = 3913.

5. Glaucoma is a disease of the eye that is manifested by high intraocular pressure. The distribution of intraocular pressure in the general population is approximately normal with mean 16 mm Hg and standard deviation 3.2 mm Hg. If the normal range for intraocular pressure is between 10 and 22 mm Hg, than what proportion of the general population would fall within this range?

ANS: Find the area between Z = 1.875 and Z = -1.875. The answer is 0.9392 (if you're close--that's fine).

6. The resting heart rate for healthy adult horses averages 46 beats per minute with a standard deviation of 8 beats per minute. A horse whose resting heart rate is in the upper 10% of the distribution of heart rates may have a secondary infection or illness that needs to be treated. How fast must a healthy horse's heart be beating to fall into this at-risk group?

ANS: The Z value with 0.10 area right of it (and 0.90 area left of it) is 1.28. The answer is 46 + 1.28 * 8 = 56.24 beats per minute.

7. GPAs of SUNY Oswego freshman biology majors have approximately the normal distribution with mean 2.87 and standard deviation .34.

a. In what range do the middle 90% of all freshman biology majors’ GPAs lie?

b. Students are thrown out of school if their GPA falls below 2.00. What proportion of all freshman biology majors are thrown out?

c. What proportion of freshman biology majors have GPA above 3.50?

ANS: a) From 2.311 to 3.429; b) 0.0052; c) 0.0322

8. The length of elephant pregnancies from conception to birth varies according to a distribution that is approximately normal with mean 525 days and standard deviation 32 days.

d. What percent of pregnancies last more than 600 days (that’s about 20 months)?

e. What percent of pregnancies last between 510 and 540 days (that’s between 17 and 18 months)?

f. How short do the shortest 10% of all pregnancies last?

ANS: d) 0.95% (a little less than 1%); e) 36.16%; f) no more than 484 days.

9. Wingspans of adult herons have approximate normal distribution with mean 125 cm and standard deviation 12 cm.

g. What proportion of herons have wingspan more than 140 cm?

h. What is the median wingspan?

i. What are the first and third quartiles of the wingspans? (25% of all herons have wingspan less than the first quartile; 25% have wingspan more than the third quartile.) Use these to obtain the IQR for heron wingspans.

ANS: g) 0.1056; h) 125 cm (the mean and median are the same for a normal distribution because of the symmetry); i) 1st quartiles – 25%; invNorm(0.25, 125, 12) = 116.91; 3rd quartiles – 75%; invNorm(0.75, 125, 12) = 133.09; IQR = Q3 – Q1 = 133.09 – 116.91 = 16.19.

10. When robins’ eggs are weighed, it turns out that they vary according to approximately the normal distribution with mean 17.5 mg and standard deviation 3.5 mg.

j. What proportion of robins’ eggs weigh between 10 mg and 25 mg?

k. How large are the largest 2% of all robins’ eggs?

ANS: j) 0.9676; k) 24.68 mg.

11. The college you are interested in gives the distribution of the SAT math scores for their students as N(580, 100). How high must a student score to be in the upper 10% of all students?

ANS: Since you want to know what the top 10% of scores would look like, the area to the left would be 0.90. That is, a z-score of z = 1.285.

[pic] giving x = 708. (or invNorm (0.9, 580, 100) )

A student would have to score 708 or higher to be in the top 10% of the student body.

12. The grading at Central High gives a B for grades between 86 and 93. On the English final for seniors, what proportion of the class would get a B if the grades were normally distributed with a mean grade of 86.34 and standard deviation of 14.23?

(a) 0.07

(b) 0.4905

(c) 0.6801

(d) 0.1896

(e) 0.0280

ANS: The correct answer is (d).

13. The mean GPA for Central High is 2.9, with the standard deviation of 0.5. Assuming the GPAs are normally distributed, what GPA score will place a student in the top 5% of the class?

(a) 3.72

(b) 3.43

(c) 2.08

(d) 2.90

(e) 3.38

ANS: The correct answer is (a).

14. On Kaylani’s last two biology exams, she scored an 87. The class mean on the first exam was 75, with a standard deviation of 8.9. The class average on the second exam was 73, with a standard deviation of 9.7. Assuming the scores on the exam were approximately normally distributed, on which exam did Kaylani score better relative the rest of her class?

(a) She scored better on the first exam.

(b) She scored better on the second exam.

(c) She scored equally on both exams.

(d) It is impossible to determine because the class sizes are unknown.

(e) It is impossible to determine because the correlation between the two sets of exam scores is not provided.

ANS: The correct answer is (b).

15. Which of the following statements is not true for normally distributed data?

(a) The mean and median are equal.

(b) The area under the curve is dependent upon the mean and standard deviation.

(c) Almost all of the data lie within three standard deviations of the mean.

(d) Approximately 68% of all of the data lies within one standard deviation of the median.

(e) When the data are normalized, the distribution has a mean μ = 0 and standard deviation σ = 1.

ANS: The correct answer is (b). The normal curve is a density curve and the area under the density curve is always 1.

16. Webb is a baseball fanatic. He keeps his own statistics on the major league teams and individual players. For the 350 regular starters, Webb has found their mean batting average is 0.229, with a standard deviation of 0.024. His sister is appalled that baseball players get paid the salaries they do and get a hit less than 25% of their attempts at bat. To further her argument, she asks for the following information:

(a) What proportion of players hit more than 25% of the times they are at bat?

(b) Since the players with the top ten batting averages get cash bonuses, what is the lowest batting average is that will receive a bonus?

ANS:

a) Hitting 25% of the time they are at bat would give the player a batting average of 0.25 or higher. [pic] Since we were asked for those whose batting average is at least 25% of their times at bat, we want the area to the right of this z-score.

So, 1 – (area to the left of z = 0.88) = 1 – 0.8106 = 0.1894. So just under 20% of the players get hits on average over 25% of the times they are at bat.

b) The top 10 players would represent [pic]or approximately 2.86% of the players. Using the Normal Distribution Table, the z-score corresponding to the top 2.86% of the players is the z-score that has 1 – 0.02857 = 0.97143 area to the left of it. That z-score is approximately 1.9. Knowing this, we can calculate the actual batting average that is 1.9 standard deviations above the mean. [pic] giving x = 0.2746. Specifically, this says that the top 10 players have batting averages of 0.2746 or higher.

17. At the end of the semester, a teacher determines the percent grade earned by students, taking into account all homework, test, and project scores. These student scores are normally distributed, with a mean of 75 and a standard deviation of 8.

(a) If a 90-80-70-60 scheme is used to determine the letter grades of students, what percent of students earned an A?

(b) If a 90-80-70-60 scheme is used to determine the letter grades of students, what percent of students earned a B?

(c) If the teacher decides to give A’s to the top 10% of the class, then what is the cut-off point for an A?

(d) If the teacher decides to give B’s to the next 20% of the class, then what is the cut-off point for a B?

ANS:

(a) In the 90-80-70-60 scheme, a grade of 90 or better earns an A.

Compute P(X ≥ 90).

normalcdf (90, 1E99, 75, 8) = 0.0304. Therefore, 3.04% of the class got A’s.

(b) In the 90-80-70-60 scheme, a grade of 80 or better but less than 90, earns a B.

Compute P(80 ≤ X < 90).

normalcdf (80, 90, 75, 8) = 0.2356. Therefore, 23.56% of the class got B’s.

(c) The teacher decides to give A’s to the top 10% of the class.

Since you want to know what the top 10% of scores would look like, the area to the left would be 0.90. That is, a z-score of z = 1.285.

[pic] giving x = 85.24. Therefore, any student with a percent grade of 85.24 or better gets an A.

(d) The teacher decides to give B’s to the next 20% of the class. This means that 30% of the students are above the cut-off point for the grade B (all A students and all B students.)

Since you want to know what the top 30% of scores would look like, the area to the left would be 0.70. That is, a z-score of z = 0.525.

[pic] giving x = 79.2. Therefore, any student with a grade of 79.2 or higher, but lower than 85.24, gets a B.

18. Railroad freight cars filled with coal have weights that are approximately normally distributed with a mean of 72 tons and a standard deviation of 9 tons. A train of 50 coal cars is compiled. The engine assigned to pull the trail can pull a maximum weight of 3,750 tons.

What percentage of all coal cars weigh less than 74 tons?

ANS:

(a) The z-score for a 74 ton car is [pic] The area under the normal curve below z = 0.22 is 0.5871. About 58.71% of cars weigh less than 74 tons.

19. Given a normal distribution with a mean of 25, what is the standard deviation if 18% of the values are above 29?

ANS: Since you want to know what the top 18% of scores would look like, the area to the left would be 0.82. That is, a z-score of z = 0.92. Thus, [pic] giving σ = 4.35.

20. Given a normal distribution with a standard deviation of 10, what is the mean if 21% of the values are below 50?

ANS: The z-score that is corresponding to 21% is -0.81. Thus, [pic] giving μ = 58.1.

21. Given a normal distribution with 80% of the values above 125 and 90% of the values above 110, what are the mean and standard deviation?

ANS: Since you want to know what the top 80% of scores would look like, the area to the left would be 0.20. That is, a z-score of z = -0.84. Thus, we have [pic]

Since you want to know what the top 90% of scores would look like, the area to the left would be 0.10. That is, a z-score of z = -1.28. Thus, we have [pic]

Now, we have -0.84σ = 125 – μ

-1.28σ = 110 – μ

Solve for μ and σ, we get μ = 153.64 and σ = 34.09.

22. Suppose that the average height of adult males in a particular locality is 70 inches with a standard deviation of 2.5 inches.

(a) If the distribution is normal, the middle 95% of males are between what two heights?

(b) Ninety percent of the heights are below what value?

(c) Ninety-nine percent of the heights are above what value?

(d) What percentage of the heights are between z-scores of ±1? Of ±2? Of ±3?

ANS:

(a) Using the 68-95-99.7 rule, the middle 95% corresponds to 2 standard deviations away from the mean, which is 70 ± 2(2.5) = 65 to 75 inches.

(b) The z-score that is corresponding 90% is 1.285. Thus, [pic] giving x = 73.205 in.

(c) Since you want to know what the top 99% of scores would look like, the area to the left would be 0.01. That is, a z-score of z = -2.325. Thus, [pic]giving x = 64.185 in.

(d) 68%, 95%, 99.7%, respectively.

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