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MATH 225N Week 5 Understanding Normal Distribution; complete solutions guide "A"

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Understanding Normal Distribution: week #5 1. Lexie averages 149 points per bowling game with a standard deviation of 14 points. Suppose Lexie's points per bowling game are normally distributed. Let... X= the number of points per bowling game. Then X∼N(149,14). Suppose Lexie scores 186 points in the game on Tuesday. The z-score when x = 186 is 2.643 - no response given. The mean is 149 - no response given. This z-score tells you that x = 186 is 2.643- no response given standard deviations to the right of the mean. The z-score can be found using this formula: z=x−μσ=186−149/14=3714≈2.643 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, scoring 186 points is 2.643 standard deviations away from the mean. A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem: μ=149 points in the game. 2. Suppose X∼N(18,2), and x=22. Find and interpret the z-score of the standardized normal random variable. This study source was downloaded by 100000831988016 from CourseHero.com on 01-19-2022 14:25:38 GMT -06:00 https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/ The z-score when x=22 is . The mean is . This z-score tells you that x=22 is standard deviations to the right of the mean. 3. Suppose X∼N(12.5,1.5), and x=11. Find and interpret the z-score of the standardized normal random variable. X is a normally distributed random variable with μ=12.5 (mean) and σ=1.5 (standard deviation). To calculate the z-score, z=x−μσ=11−12.51.5=−1.51.5=−1 This means that x=11 is one standard deviation (1σ) below or to the left of the mean. This makes sense because the standard deviation is 1.5. So, one standard deviation would be (1)(1.5)=1.5, which is the distance between the mean (μ=12.5) and the value of x (11). 2 18 2 This study source was downloaded by 100000831988016 from CourseHero.com on 01-19-2022 14:25:38 GMT -06:00 https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/ 5. Isabella averages 17 points per basketball game with a standard deviation of 4 points. Suppose Isabella's points per basketball game are normally distributed. Let X= the number of points per basketball game. Then X∼N(17,4). 3,17,3 6. Suppose X∼N(13.5,1.5), and x=9. Find and interpret the z-score of the standardized normal random variable.-3,13.5,3 7. Suppose X∼N(10,0.5), and x=11.5. Find and interpret the z-score of the standardized normal random variable. 3,10,3 This means that x=11.5 is three standard deviations (3σ) above or to the right of the mean. This makes sense because the standard deviation is 0.5. So, three standard deviations would be (3)(0.5)=1.5, which is the distance between the mean (μ=10) and the value of x (11.5). 8. Annie averages 23 points per basketball game with a standard deviation of 4 points. Suppose Annie's points per basketball game are normally distributed. Let X= the number of points per basketball game. Then X∼N(23,4). 2.75,23,2.75 9. Suppose X∼N(9,1.5), and x=13.5. Find and interpret the z-score of the standardized normal random variable. 3,9,3 10. Rosetta averages 148 points per bowling game with a standard deviation of 14 points. Suppose Rosetta's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(148,14). 2.714,148,2.714 This study source was downloaded by 100000831988016 from CourseHero.com on 01-19-2022 14:25:38 GMT -06:00 https://www.coursehero.com/file/58202586/Understanding-Normal-Distribution-week-5docx/ 11. Suppose X∼N(5.5,2), and x=7.5. This means that x=7.5 is one standard deviation (1σ) above or to the right of the mean, μ=5.5. 12. Jerome averages 16 points a game with a standard deviation of 4 points. Suppose Jerome's points per game are normally distributed. Let X = the number of points per game. Then X∼N(16,4). Suppose Jerome scores 10 points in the game on Monday. The z-score when x=10 is _______. This z-score tells you that x=10 is _______ standard deviations to the _______(right /left) of the mean, _______. [Show More]

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Sugar canes have lengths, X, that are normally distributed with mean 365.45 centimeters and standard deviation 4.9 centimeters. What is the probability of the length of a randomly selected cane bein...

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