Difference between revisions of "EBook Problems Normal Critical"
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:''(e) Relative to other students, Wanda did better in calculus. | :''(e) Relative to other students, Wanda did better in calculus. | ||
+ | {{hidden|Answer|(b)}} | ||
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+ | ===Problem 16=== | ||
+ | Suppose that the heights of college women are nearly normal with mean = 65 inches and deviation = 2.0 inches. If talented gymnast Vanessa is at the 10th percentile in height for college women, determine Vanessa's height. | ||
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+ | *Choose one answer. | ||
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+ | :''(a) 60.50 inches | ||
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+ | :''(b) 62.44 inches | ||
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+ | :''(c) 62.00 inches | ||
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+ | :''(d) 64.50 inches | ||
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+ | :''(e) Cannot be determined without the data | ||
{{hidden|Answer|(b)}} | {{hidden|Answer|(b)}} | ||
Revision as of 19:09, 8 January 2009
EBook Problems Set - Nonstandard Normal Distribution: Finding Scores (Critical Values)
Problem 1
Given the following summary statitics, which of the statements below is true?
Chemistry test score of Class A: Mean = 70 Standard deviation = 10 Chemistry test score of Class B: Mean = 70 Standard deviation = 5
Jack is in class A and got a score of 80 John is in class B and got a score of 80
- Choose one answer.
- (a) John's performance in chemistry is better than Jack's.
- (b) Jack and John did equally well in chemistry.
- (c) John's and Jack's scores in chemistry are not comparable.
- (d) We need more information to compare Jack's and John's scores.
Problem 2
A statistics professor teaches two different sections of the same class. To prevent cheating, the classes get different midterms. This year, the midterm for Section A had a mean of 60% with a standard deviation of 10%. The midterm for Section B had a mean of 65% and a standard deviation of 5%. Kefei scored 75% in Section A, while her friend Bob in Section B scored 75%. Grades on this midterm are based upon how each student does relative to the other students in his or her class.
Which of the following best describes the situation?
- Choose one answer.
- (a) Because the midterms contained different questions, we can't compare their performance.
- (b) Relative to other classmates, both Bob and Kefei did equally well.
- (c) Relative to other classmates, Bod did better on the exam than Kefei.
- (d) Relative to other classmates, Kefei did better on the exam than Bob.
Problem 3
In Japan there is an annual turkey dog eating contest. The number of turkey dogs that contestants eat are normally distributed with a mean of 36 turkey dogs and a standard deviation of 6 turkey dogs. A contestant eats 27 turkey dogs. What is his z-score?
- Choose one answer.
- (a) 9
- (b) 1.5
- (c) 6
- (d) -9
- (e) -1.5
Problem 4
Volunteers from Heal the Bay determined that the lengths of individual shellfish in the population of 10,000 shellfish are nearly normal with mean 10 centimeters and deviation of 0.2 centimeters. Determine the shortest interval that contains approximately 4,000 shellfish.
- Choose one answer.
- (a) 9.744cm to 10cm
- (b) 9.928cm to 10.080cm
- (c) 0cm to 9.949cm
- (d) 9.774cm to 10.256cm
- (e) 9.895cm to 10.105cm
Problem 5
You read that the heights of college women are nearly normal with a mean of 65 inches and a standard deviation of 2 inches. If Vanessa is at the 10th percentile (shortest 10% for women) in height for college women, then her height is closest to:
- Choose one answer.
- (a) 64.5 inches
- (b) It cannot be determined from this information
- (c) 60.5 inches
- (d) 62.44 inches
Problem 6
The length of human pregnancies from conception to birth has a distribution that is approximately normal, with a mean 266 days and a standard deviation of 16 days. How short are the shortest 4% of pregnancies?
- Choose one answer.
- (a) Less than 210 days
- (b) Less than 266 days
- (c) Less than 238 days
- (d) Less than 280 days
Problem 7
In standardized scores of achievement motivation that are normally distributed, the mean score is 35 with a standard deviation of 14. Higher scores correspond to more achievement motivation.
Shamu scored in the top 20% of the test takers. What was her score?
- Choose one answer.
- (a) 20
- (b) 23
- (c) 80
- (d) 47
Problem 8
Avoiding an accident while driving can depend upon reaction time. That time, measured from the moment the driver first see the danger until he or she gets a foot on the brake pedal, is thought to follow a Normal Model with mean = 1.50 seconds and standard deviation = 0.18 seconds. About what percent of the drivers have a reaction time more than 1.25 seconds?
- Choose one answer.
- (a) None of these values
- (b) 8.2%
- (c) 1.39%
- (d) 91.8%
Problem 9
Among first year students at a certain university, scores on the verbal SAT follow the normal curve. The average is around 500 and the SD is about 100. Tatiana took the SAT, and placed at the 85% percentile. What was her verbal SAT score?
- Choose one answer.
- (a) 90
- (b) 403
- (c) 560
- (d) 604
Problem 10
There is a Turkey Dog eating contest in Japan. The number of dogs eaten by contestants are normally distributed with mean = 36 and standard deviation = 6). The great Kobiyashi eats an amazing 54 turkey dogs. What percentage of contestants eat fewer turkey dogs than Kobiyashi?
- Choose one answer.
- (a) 99.87%. Practically nobody can out eat Kobiyashi
- (b) 95%. Because of the golden rule
- (c) 0.13%. Kobiyashi should try a different sport
- (d) 100%. This competition was over before it started
Problem 11
Again, we are at a turkey dog eating contest in Japan (number of dogs eaten are normally distributed with a mean of 36 and a sd of 6). The lowest 10% of the contestants eat fewer than how many turkey dogs?
- Choose one answer.
- (a) 29.32 dogs
- (b) 33 dogs
- (c) 10 dogs
- (d) 25 dogs
Problem 12
Comparing Z-scores
Which of the following statements about z-scores is true?
A: yA = 25.4 with mean = 12.9 and deviation = 3.7 or B: yB = 137.5 with mean = 72.7 and deviation = 21.1
- Choose one answer.
- (a) ZA is larger
- (b) We cannot compare z-scores since the distributions are not normal
- (c) We cannot compare z-scores since we do not know the sample sizes
- (d) ZB is larger
Problem 13
Suppose that the mean height of the 35 students in your discussion section was 68.5 inches with a standard deviation of 2.4 inches when a new student, who was 6 feet tall, joined your section. Now calculate the mean for all 36 students.
- Choose one answer.
- (a) 68.70 inches
- (b) 74.5 inches
- (c) 68.60 inches
- (d) 70.25 inches
Problem 14
During a recent physical, Gillian's doctor told her that the standardized score (z-score) for her systolic blood pressure, as compared to the blood pressure of other women her age, was 1.50. Which of the following is the best interpretation of this standardized score?
- Choose one answer.
- (a) Gillian's systolic blood pressure is 150
- (b) Gillian's blood pressure is 1.50 above the mean systolic blood pressure for women her age
- (c) Gillian's systolic blood pressure is 1.50 times the mean systolic blood pressure for women her age.
- (d) Gillian's systolic blood pressure is 1.50 standard deviations above the mean systolic blood pressure for women her age.
- (e) Only 1.5% of women who are Gillian's age have a higher systolic blood pressure.
Problem 15
Suppose that the scores on the chemistry final exam are nearly normal with mean of 75 and standard deviation of 12. The scores of the calculus final are also nearly normal with mean of 80 and deviation of 8. Wanda scores 81 on the chemistry exam and 84 on calculus final. Relative to other students in each course, how did Wanda compare?
- Choose one answer.
- (a) There is not enough information for comparison, because the number of students in each class is not known.
- (b) Relative to other students, Wanda did equally well on both exams.
- (c) Relative to other students, Wanda did better in chemistry.
- (d) There is no basis for comparison, since the subjects are different and in different departments.
- (e) Relative to other students, Wanda did better in calculus.
Problem 16
Suppose that the heights of college women are nearly normal with mean = 65 inches and deviation = 2.0 inches. If talented gymnast Vanessa is at the 10th percentile in height for college women, determine Vanessa's height.
- Choose one answer.
- (a) 60.50 inches
- (b) 62.44 inches
- (c) 62.00 inches
- (d) 64.50 inches
- (e) Cannot be determined without the data
- Back to Ebook
- SOCR Home page: http://www.socr.ucla.edu
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