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5. On pages 11 and 12 of the chapter 5 notes there are some problems that were in the chapter 4 but they are repeated in chapter 5 because the probabilities of the z-scores can now be found. What is the correctly rounded probability of the CIA problem for 43 or fewer females in the problem from page 11? 0.0014 0.0015 0.9985 0.9986

5. On pages 11 and 12 of the chapter 5 notes there are some problems that were in the chapter 4 but they are repeated in chapter 5 because the probabilities of the z-scores can now be found. What is the correctly rounded probability of the CIA problem for 43 or fewer females in the problem from page 11? 0.0014 0.0015 0.9985 0.9986

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Hi, I attached pages 11 and 12 in my notes, for reference to the question. 

Any help appriciated as soon as possible, thanks.

5. On pages 11 and 12 of the chapter 5 notes there are some problems that were in the chapter 4 but they are repeated in chapter 5 because the probabilities of the z-scores can now be found. What is the correctly rounded probability of the CIA problem for 43 or fewer females in the problem from page 11? 0.0014 0.0015 0.9985 0.9986
Transcribed Image Text:5. On pages 11 and 12 of the chapter 5 notes there are some problems that were in the chapter 4 but they are repeated in chapter 5 because the probabilities of the z-scores can now be found. What is the correctly rounded probability of the CIA problem for 43 or fewer females in the problem from page 11? 0.0014 0.0015 0.9985 0.9986
Example If we apply these formulas to our CIA example, we see that the mean number of females expected in a sample of 5 randomly selected people is μ = np = 5.0.498 = 2.49, while the standard deviation is: 0 = |npq = V5(0.498)(0.502) = 1.118025 We ask whether observing x ≥ 4 females in a random sample of five people would be an unusually high number of females. The Z-score can answer this, so we compute Z = Now let's convert this z-score to a probability. x-μ 3.5-2.49 O 1.118 Z = We discussed in chapter 4 why we must use x = 3.5 instead of using x = 4. We will discuss this issue again in this chapter. - Question: from the CIA example, would 43 or fewer females in a random group of 120 people be unusual? To answer this, we simply compute the z-score, 0.903. x-μ O 43.5 - 59.76 5.47718 which tells us that the cutoff value of 43.5 is -2.969 standard deviations below average. Again, for this chapter, we can convert this z-score to a probability. 11 = -2.969 We just used the standard deviation method to determine if a value is highly unusual (or an outlier). It is better to use the estimation method in this case since the exact method would require us to use the exact formula (page 10) for every value of x between 0 and 43. Example Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such candies is randomly selected. (a) Find the mean and standard deviation for the number of yellow candies in such groups of 100. (b) A random sample of 100 plain M&M candies is taken, and of these, 8 are yellow. Does this result seem to indicate that the Mars, Inc. claimed rate of 14% is wrong? 12
Transcribed Image Text:Example If we apply these formulas to our CIA example, we see that the mean number of females expected in a sample of 5 randomly selected people is μ = np = 5.0.498 = 2.49, while the standard deviation is: 0 = |npq = V5(0.498)(0.502) = 1.118025 We ask whether observing x ≥ 4 females in a random sample of five people would be an unusually high number of females. The Z-score can answer this, so we compute Z = Now let's convert this z-score to a probability. x-μ 3.5-2.49 O 1.118 Z = We discussed in chapter 4 why we must use x = 3.5 instead of using x = 4. We will discuss this issue again in this chapter. - Question: from the CIA example, would 43 or fewer females in a random group of 120 people be unusual? To answer this, we simply compute the z-score, 0.903. x-μ O 43.5 - 59.76 5.47718 which tells us that the cutoff value of 43.5 is -2.969 standard deviations below average. Again, for this chapter, we can convert this z-score to a probability. 11 = -2.969 We just used the standard deviation method to determine if a value is highly unusual (or an outlier). It is better to use the estimation method in this case since the exact method would require us to use the exact formula (page 10) for every value of x between 0 and 43. Example Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such candies is randomly selected. (a) Find the mean and standard deviation for the number of yellow candies in such groups of 100. (b) A random sample of 100 plain M&M candies is taken, and of these, 8 are yellow. Does this result seem to indicate that the Mars, Inc. claimed rate of 14% is wrong? 12
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