As always, show your work. 1) IO scores are normally distributed with a mean of 100 and a standard deviation of 15. Nine people are randomly selected. What is the probability that the average of their 9 IO scores is less than 110? 2) Men's weights are normally distributed with a mean of 189le and a standard deviation of 39le. Four men are randomly selected. What is the probability that the average of their 4 weights is more than 195lbs? 3) Women's heights are normally distributed with a mean of 63.7 inches and a standard deviation of 2.6 inches. Seven women are randomly selected. What is the probability that the average of their 7 heights is more than 66 inches? FOR REVIEW: 4) Women's weights are normally distributed with a mean of 171lbs and a standard deviation of 46le. If one woman is randomly selected, what is the probability that her weight is less than 150lbs? 5) Mensa International calls itself \"the international high IQ society,\" and has more than 100,000 members. Membership requires an IQ score over 130.5. If one person is randomly selected, what is the probability that helshe is eligible for Mensa membership? Lesson 3 Chapter 6 In the previous lesson we looked at probabilities of one randomly selected data point: one random IQ score, one random man's height, one pregnancy length. In this lesson we will consider probabilities of randomly selected groups of data. As a review and for reference, let's find the probability of one randomly selected person having and IQ of less than 90. Step 1: Convert the 90 to a standard normal z. 351\" Z= x=90 ,u.=100 6:15 0' 90100 _ 1 Z = = -0.67 15 15 Step 2: Take the z of -0.67 to the table and nd the probability of 0.2514. Step 3: Since the table values are always from the left tail up, 0.2514 is the probability of less than 90. Conclusion: The probability that one randomly selected person has an IQ under 90 is 0.2514, 25%. Now we consider a slightly different question. If a group of FOUR people is randomly selected, what is the probability that the average of their four IQ scores is under 90? We use the same procedure as above, with one slight change to the zscore converter. When selecting a group (in this case 4) instead ofjust one, we must divide the standard deviation by the square root of the sample size n, in this case n = 4. x-# Z= a/ Note the only difference from our previous converter (below) is the division of (I by . EXAMPLE 1: If 4 people are randomly selected, find the probability that the average of their 4 IQ scores is less than 90. Step 1: Take note that we have a sample size of 4, n = 4, then convert the 90 to a standard normal 2. 95.\" ZZJ/ x=90 #:100 0:15 n=4 90100 _ 90100 _ 10 : ls/JZ _ 15/2 _ E = '1'33 Step 2: Take the z of -1.33 to the table and find the probability of 0.0918. Step 3: Since the table values are always from the left tail up, 0.0918 is the probability of less than 90. Conclusion: The probability that a group of 4 randomly selected people has an average IQ under 90 is 0.0918, 9.2%. EXAMPLE 2: If 16 men are randomly selected, find the probability that their average height is more than 6 feet (72 inches). Recall that for men's height, the mean is 68.6\" and the standard deviation is 2.8\". Step 1: Take note that we have a sample size of 16, n = 16, then convert the 7'2 to a standard normal 2. 95.\" ZZJ/ x=72 ,u=68.6 0:2.8 n=16 7268.6 _ 7268.6 _ 3.4 = ash/E _ 2.8/4 _ E = 4'86 Step 2: Take the z of 4.86 to the table and find the probability of 0.9999. Step 3: Since the table values are always from the left tail up, 0.9999 is the probability of less than 72. We want 1 - 0.9999 = 0.0001, the probability of more than 72. Conclusion: The probability that a group of 16 randomly selected men has an average height over 6 feet is 0.0001, 0.01%. Consider the true and tragic story of the Ethan Allen Tour Boat that sank in Lake George in New York state in October of 2005. 20 of the 47 passengers drowned. Sadly, the maximum occupancy for the boat was still based on norms from 1960. On average, people were heavier in 2005 than in 1960. For the 1960 maximum occupancy calculations, an average weight of 140|bs per person was assumed. That allowed for a maximum of 50 passengers. After the boat sank, average weight per person was updated to 174le. The boat was rated for a passenger weight capacity of 2,436 lbs. This allowed for a maximum occupancy of 14 passengers. 14 passengers, times 174 pounds per person equals 2,436 pounds. EXAMPLE 3: On average, men are heavier than women. We will assume the worstcase scenario of 14 men aboard the boat. Men's weights have a mean of 189le and a standard deviation of 39lbs. An average of 174 pounds is allowed for each passenger. So if the average weight of the 14 men is above 174 pounds, the weight capacity of the boat will be exceeded. If 14 men are randomly selected, what is the probability that their average weight will exceed 174 pounds? Step 1: Take note that we have a sample size of 14, n = 14, then convert the 174 to a standard normal 2. x ZZJ/ x=174 u=189 6:39 n=14 174189 _ 174189 _ 15 = 39/x/ _ 39/374 _ 10.43 = '1 '44 Step 2: Take the z of -1.44 to the table and find the probability of 0.0749. Step 3: Since the table values are always from the left tail up, 0.0749 is the probability of less than 174. We want 1 0.0749 = 0.9251, the probability of more than 174. Conclusion: The probability that a group of 14 randomly selected men has an average weight over 174 pound is 0.9251, 93%. There is a 93% chance that if there are 14 men aboard the boat, the weight capacity will be exceeded. Perhaps the allowance of 174 pounds per person is still too low