Homework: Week 5 Normal Distributions Score: 0 of 1 pt 3 of 15 (11 complete HW S
ID: 3176538 • Letter: H
Question
Homework: Week 5 Normal Distributions Score: 0 of 1 pt 3 of 15 (11 complete HW Score: 41.11%, 6.17 of 15 pts 5.1.33 Question Help The mean weight of a breed of yearing cattle is 1079 pounds. Suppose that weights of all such animals can be described by a Normal model with a standard deviation of 65 pounds. What weight would be considered unusually low for such an animal? Select the correct choice below and fill in the answer boxes within your choice. O A Any weight more than 3 standard deviations below the mean, or less than pounds, is unusually low. one would expect to see a steer 2 standard deviations below the mean, or less than pounds only rarely. O B. Any weight more than 2 standard deviations below the mean, or less than pounds, is u low. One would expect to see a steer 3 standard deviations below the mean, or less than pounds only rarely. nusually O C. Any weight more than 1 standard deviations below the mean, or less than pounds, is unusually low. One would expect to see a steer 2 standard deviations below the mean, or less than pounds only rarely.Explanation / Answer
Q. 5.1.33 Mean weight = 1079 pounds and standerd deviation = 65 pounds
so any weight which is more than 2 standerd deviation below the mean or less than (1079 - 2 * 65 = 949 pounds), is unusually low.one would expect to see a steer 3 standard deviation below the mean, or less than (1079 - 3 * 65 = 884 pounds),only rarely. Option B is the correct option.
Section 5.1 exercise 3
January Average Tempp = 36 C and standard deviation of temp. = 11C
January Average Tempp = 74 C and standard deviation of temp. = 7C
so P(T = 56) in any of 2 months can be calculated by Z value for both temperature distribution is
for january Z value = (56 - 36)/11 = 1/82
for July Z value = (56 -74)/7 = - 2.57
So, option B is correct. It is more unusual to have a day with a high temperature of 56C in July. A high temperature in 56C in July is 2.57 standerd deviations below the mean and a high temperature of 56C is only 1.82 times standerd deviations above the mean.
Example 5.1.21
First stats exam has mean = 60 and s.d. = 10 marks
second stats exam has mean = 80 and s.d. = 5 marks
derrick scored 80 + 80 in both tests and Julie got 70+ 90 in both tests.
Julie claimed she performed better.
Option C is correct.Julie's claim is incorrect. They both totaled 160 points in the two exams so neither student did better than the other. We can't add two Z -values for different samples to proove which one is good, so option A and option B both are ncorrect.
By the way option A is "Julie's claim is correct.Derrick's Z -score are [(80-60)/10 = 2] for the first test and 0 for the second test.Julie's Z-score are [(70 -60)/10 =1] for the first test and [(90-80)/5 = 2] for the second test.Derrick's total is 2, which is less than Julie's total 3."
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