Studocu Worksheet-13-normal Distribution-1 PDF

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Math130

WORKSHEET 13 Normal Distribution (Use standard normal distribution table or TI 83, 84 Calculator)

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NORMAL PROBABILITY DENSITY CURVES: 1. Shape: bell 2. Domain: - ∞ to ∞ 3. Parameters required: Mean and standard deviation 4. Skewness: 0 5. Probability:

STANDARD NORMAL PROBABILITY DENSITY CURVES: 1. Shape: bell 2. Domain: - ∞ to ∞ 3. Parameters required: None 4. Skewness: 0 5. Probability:

Q1: Suppose the random variable Z follows standard normal distribution. Values obtained from a Z table (a) How likely is it to see Z>0? = 0.5000 (b) How likely is it to see Z -0.9)) Z > -0.9 From table P (Z= -0.9) = 0.1841 P (Heavier than John) = 1-0.1841 = 0.8159 (iv) What is the probability that a randomly chosen male adult is heavier than Peter? P(X>192) P(X>192) = P((X-168)/20> (192-168)/2) P(X>192) = (z > 1.2)) Z > 1.2 From table P (Z= 1.2) = 0.8849 1-0.8849= 0.1151 (v) Jack is small, only 10% of the male adults are lighter than Jack. What is the z-score of Jack’s weights? How much does Jack weigh? P(Jack) =10/100 = 0.1 P (Z= x) = 0.1 Z scores with 0.10 are to its left is Z=-1.28

Z = (X- μ)/ σ -1.28 = (X- 168)/ 20 Solving for X = 142.4 The weight of Jack is 142.4lbs and Z-score is -1.28 (vi) Dan is a big guy; he is heavier than two-thirds of the male adults. Find the z-score of Dan’s weight and Dan’s actual weight. Z = (X- μ)/ σ 0.43 = (X- 168)/ 20 Solving for X = 176.6 The weight of Dan is 176.6lbs and Z-score is 0.43 2. Professor Smith teaches a large English class. After grading the final exam, Professor Smith decided that the final grades are approximately normal distributed with mean of 135 and standard deviation of 16. Professor Smith decided to give the top 5% of the students ‘A’, next 20% students ‘B’, next 30% students ‘C’, the bottom 10% students ‘F’, and the rest students ‘D’. Let a, b, c, d be the cutoff points for grades ‘A’, ‘B’, ‘C’ and ‘D’. μ = 135 σ = 16 (i) Determine the z-scores of the cutoff points, za, zb, zc and zd. P (Za)=0.05 From table, Z score for P= 0.05 za = -1.645 P(x>b) =0.25 P[(x-135)/16 >(b-135)/16] =0.25 Z [(b-135)/16] = 0.75 Z [(b-135)/16] = Z (0.67) zb = 0.67 P(zc) = 0.55 P[(x-135)/16...