1.11 lecture13

1.11 lecture13 - 1 STAT 1301 Probability &...

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Unformatted text preview: 1 STAT 1301 Probability & Statistics Lecture 13 Lecture 13 Standardization • (100 p )th percentile for N ( , 2 ) = + · (100 p )th percentile for N (0, 1). • So if X ~ N ( , 2 ), then X = + · Z . Markov Inequality • If X is a positive random variable with finite mean, then for any constant c > 0 , Chebyshev’s Inequality • If the random variable X has a finite mean and finite variance σ 2 , then for any constant k > 0, • The probability that X deviates from the mean more than 2 standard deviations is at most 0.25. Proof • By Markov inequality Standard Normal Table • If X is normal, then – about 68% of the x 's are within 1 SD of the mean; – about 95% of the x 's are within 2 SDs of the mean; – about 99.7% of the x 's are within 3 SDs of the mean. 2 Example • The lengths of fish in a certain fish population follows a normal distribution with = 54 mm and = 4.5 mm. – What percentage of the fish are 50 to 60 mm long? * Let Z = ( X- )/ . Then z 1 =(50 - 54)/4.5= -.89, z 2 =(60 - 54)/4.5=1.33. P (50 X 60)= P 89 Z 1 33) = 9082 1867= 7215 P (50 X 60)= P (-.89 Z 1.33) =.9082 - .1867= .7215. – What is the 70th percentile of the fish length? What is the 90th percentile? * From the standard normal Table , Z .70 = 0.52. So, X .70 = 54 + 4.5 · 0.52 = 56.3 * Similarly, Z .90 = 1.29. and X .90 = 54 + 4.5 ·1.29 = 59.80....
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1.11 lecture13 - 1 STAT 1301 Probability &...

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