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North Carolina State University STAT 370: Probabilityy and Statistics for Engineers [Section 002] Instructor: Hua Zhou Harrelson Hall 210 11:45AM-1:00PM, Apr 11, 2012 Midterm 2 Announcements • Midterm 2 returned • Quiz 4 returned • HW 9 (continuous R.V., 10pt) due Friday Apr 13 @ 11:59PM • HW 10 (normal distribution, 10pt) due Friday Apr 13 @ 11:59PM • HW 11 (sampling dist and C.I., 10pt) due Apr 20 @ 11:59PM • HW 12 (hypothesis testing, testing 10pt) due Apr 27 @ 11:59PM Plan • Last time: Continuous random variable, normal distribution, standard normal table • Today: Normal probability calculation, start the last topic (inferential statistics) 1 The Standard Normal Table • The standard normal table is a table of areas under the standard normal density curve. The table entry for each value z is the area under the curve to the left of z. Normal table gives us CDF values (P(Z<=z)) Standard Normal Distribution • Here are some examples of probability calculations from N(0,1): The 68-95-99.7 Rule for Normal distribution Pr[ 1 Z 1] .6826 Pr[ 2 Z 2] .9546 P r[ 3 Z 3] .9974 This is where the 68-95-99.7 empirical rule comes from. 2 Means/Variances Under Linear Transformation Non-standard Normal variables: calculation of Pr(a<X<b) • Consider X~ N(µ,σ2), how to calculate probability involving X? • Question: What is the distribution of Z= (X-µ)/σ ? Answer: Z is a standard normal. This is because: 1) Linear transformations applied to normal variables preserve normality (e.g. Y= aX+b, then if X is normal so i Y) is Y). 2) Remember common rules for means and variances: • If X is a RV and a and b are constants, then 1. 2 3. E ( X a) E ( X ) a E (aX ) aE ( X ) E (aX b) aE ( X ) b Var ( X a) Var ( X ) 1. 2. Var (aX ) a 2Var ( X ) 3 Var (aX b) a 2Var ( X ) Non-standard Normal variables: calculation of Pr(a<X<b) Example: 1. Assume X is normal with µ=30 µ 30 and σ2=25. 25. Determine Pr[X <= 35]: Solution: i. Convert x-value to a z-score xvalue mean 35 30 =1.00 = z-score std .deviation 5 Thi tells This t ll us how h many standard t d d deviations d i ti ((and d which hi h direction) away from the mean the x-value is. ii. P[X<= 35] = Pr[Z <= 1.00] = 0.8413 [using the Tablle] In class exercise • Let X N(6,5. 2 ) . Find: • a) P(6 X 12) P( X 12) P( X 6) ..... • b) P( X 21) 1 P( X 21) ...... • c) P( X 6 5) P(5 X 6 5) ..... 3 In class exercise [solutions] Example: Young Women’s Height • The heights of young women are approximately normal with mean = 64.5 inches and std.dev. = 2.5 inches. • Let X N(6,5. 2 ) . Find: • a) P(6 X 12) P( X 12) P( X 6) 12 6 66 ) P(Z ) P(Z 6/ 5) P(Z 0) 5 5 0.8849 0.5 0.3849 P(Z • b) P( X 21) 1 P( X 21) 1 P(Z 21 6 ) 5 1 P(Z 3) 1 .9986 .0014 • c) P( X 6 5) P(5 X 6 5) P(1 X 11) P( X 11) P( X 1) P(Z 1) P(Z 1) 0.8413 0.1586 0.6827 Example: Young Women’s Height (cont’d) The heights of young women are approximately normal with mean = 64.5 inches and std.dev. = 2.5 inches. (a) Determine the % of young women between 62 and 67 ? (b) Determine the % of young women lower than 62 or taller than 67? (c) Determine the % between 59.5 and 62? (d) Determine the % taller than 68.25? 14 In class exercise • SAT verbal scores are known to be normally distributed with mean µ = 505 and standard deviation σ = 110 based on data from the College Board. (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of test takers who scored less than 395. (c) The area under the normal curve to the left of X=395 is 0 0.1587. 1587 Provide two interpretations of this result result. (The probability that a randomly selected person scores less than 395, and the proportion of the population that scores less than 395.) (d) Determine the 95th percentile of this distribution. Begin with 1.645 from Z. 4 In class exercise (cont’d) • SAT verbal scores are known to be normally distributed with mean µ = 505 and standard deviation σ = 110 based on data from the College Board. (e) Determine the values that determine the middle 95% of the scores (by middle we refer to an interval that is symmetric about the mean) In class exercise The heights of a pediatrician’s 200 three-year-old females are normally distributed with mean 38.72 inches and standard deviation 3.17 inches. The pediatrician wishes to determine the middle 98% of heights (thus we want the 1st and 99th percentile). What are these values? In class exercise The diameters of bearing journals ground on a particular grinder can be described as normally distributed with mean 2 005 in. 2.005 in and standard deviation 0 0.005 005 in. in a) If engineering specifications on these diameters are 2.000in. 0.005 in., what fraction of these journals are in specifications? (b) Assume that journal diameters are independent of other journals Four journals are chosen at random journals. random, and let Y be the number of these journals that are within specifications. Determine the probability mass function of Y, and the probability that at least one of the four journals is within specifications. 5