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Normal Distribution Normal Distribution Table A.3 z ··· -1.2 -1.1 ··· 1.6 1.7 ··· Standard Normal Curve Areas .00 ··· 0.1151 0.1357 ··· 0.9452 0.9554 ··· .01 ··· 0.1131 0.1335 ··· 0.9463 0.9564 ··· .02 ··· 0.1112 0.1314 ··· 0.9474 0.9573 ··· .03 ··· 0.1094 0.1292 ··· 0.9484 0.9582 ··· .04 ··· 0.1075 0.1271 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· ··· ··· ··· .09 ··· 0.0985 0.1170 ··· 0.9545 0.9633 ··· Normal Distribution Normal Distribution Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and (c)P(−1.12 < Z ≤ 1.61). Normal Distribution Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and (c)P(−1.12 < Z ≤ 1.61). z ··· -1.2 -1.1 ··· 1.6 1.7 ··· .00 ··· 0.1151 0.1357 ··· 0.9452 0.9554 ··· .01 ··· 0.1131 0.1335 ··· 0.9463 0.9564 ··· .02 ··· 0.1112 0.1314 ··· 0.9474 0.9573 ··· .03 ··· 0.1094 0.1292 ··· 0.9484 0.9582 ··· .04 ··· 0.1075 0.1271 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· ··· ··· ··· .09 ··· 0.0985 0.1170 ··· 0.9545 0.9633 ··· Normal Distribution Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and (c)P(−1.12 < Z ≤ 1.61). z ··· -1.2 -1.1 ··· 1.6 1.7 ··· .00 ··· 0.1151 0.1357 ··· 0.9452 0.9554 ··· .01 ··· 0.1131 0.1335 ··· 0.9463 0.9564 ··· P(Z ≤ 1.61) = 0.9463; .02 ··· 0.1112 0.1314 ··· 0.9474 0.9573 ··· .03 ··· 0.1094 0.1292 ··· 0.9484 0.9582 ··· .04 ··· 0.1075 0.1271 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· ··· ··· ··· .09 ··· 0.0985 0.1170 ··· 0.9545 0.9633 ··· Normal Distribution Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and (c)P(−1.12 < Z ≤ 1.61). z ··· -1.2 -1.1 ··· 1.6 1.7 ··· .00 ··· 0.1151 0.1357 ··· 0.9452 0.9554 ··· .01 ··· 0.1131 0.1335 ··· 0.9463 0.9564 ··· .02 ··· 0.1112 0.1314 ··· 0.9474 0.9573 ··· .03 ··· 0.1094 0.1292 ··· 0.9484 0.9582 ··· .04 ··· 0.1075 0.1271 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· ··· ··· ··· .09 ··· 0.0985 0.1170 ··· 0.9545 0.9633 ··· P(Z ≤ 1.61) = 0.9463; P(Z > −1.12) = 1 − P(Z ≤ −1.12) = 1 − 0.1314 = 0.8686; Normal Distribution Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and (c)P(−1.12 < Z ≤ 1.61). z ··· -1.2 -1.1 ··· 1.6 1.7 ··· .00 ··· 0.1151 0.1357 ··· 0.9452 0.9554 ··· .01 ··· 0.1131 0.1335 ··· 0.9463 0.9564 ··· .02 ··· 0.1112 0.1314 ··· 0.9474 0.9573 ··· .03 ··· 0.1094 0.1292 ··· 0.9484 0.9582 ··· .04 ··· 0.1075 0.1271 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· ··· ··· ··· .09 ··· 0.0985 0.1170 ··· 0.9545 0.9633 ··· P(Z ≤ 1.61) = 0.9463; P(Z > −1.12) = 1 − P(Z ≤ −1.12) = 1 − 0.1314 = 0.8686; P(−1.12 < Z ≤ 1.61) = P(Z ≤ 1.61) − P(Z ≤ −1.12) = 0.9463 − 0.1314 = 0.8149. Normal Distribution Normal Distribution Many tables for the normal distribution contain only the nonnegative part. Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z .00 .01 .02 ··· ··· ··· ··· 1.6 0.9452 0.9463 0.9474 1.7 0.9554 0.9564 0.9573 ··· ··· ··· ··· What is P(Z < −1.63)? .03 ··· 0.9484 0.9582 ··· .04 ··· 0.9495 0.9591 ··· ··· ··· ··· ··· ··· .09 ··· 0.9545 0.9633 ··· Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z .00 .01 .02 .03 .04 ··· .09 ··· ··· ··· ··· ··· ··· ··· ··· 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 0.9633 ··· ··· ··· ··· ··· ··· ··· ··· What is P(Z < −1.63)? By symmetry of the pdf of Z , we know that P(Z < −1.63) = P(Z > 1.63) = 1 − P(Z ≤ 1.63) = 1 − 0.9484 = 0.0516 Normal Distribution Many tables for the normal distribution contain only the nonnegative part. z .00 .01 .02 .03 .04 ··· .09 ··· ··· ··· ··· ··· ··· ··· ··· 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 0.9633 ··· ··· ··· ··· ··· ··· ··· ··· What is P(Z < −1.63)? By symmetry of the pdf of Z , we know that P(Z < −1.63) = P(Z > 1.63) = 1 − P(Z ≤ 1.63) = 1 − 0.9484 = 0.0516 Normal Distribution Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X , η(p), is defined by Z η(p) p = F (η(p)) = f (y )dy −∞ Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X , η(p), is defined by Z η(p) p = F (η(p)) = f (y )dy −∞ Similarly, the (100p)th percentile of the standard normal rv Z is defined by Z η(p) 1 2 √ e −y /2 dy p = F (η(p)) = 2π −∞ Normal Distribution Recall: The (100p)th percentile of the distribution of a continuous rv X , η(p), is defined by Z η(p) p = F (η(p)) = f (y )dy −∞ Similarly, the (100p)th percentile of the standard normal rv Z is defined by Z η(p) 1 2 √ e −y /2 dy p = F (η(p)) = 2π −∞ We need to use the table for normal distribution to find (100p)th percentile. Normal Distribution Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z ··· 1.6 1.7 ··· .00 ··· 0.9452 0.9554 ··· .01 ··· 0.9463 0.9564 ··· .02 ··· 0.9474 0.9573 ··· .03 ··· 0.9484 0.9582 ··· .04 ··· 0.9495 0.9591 ··· 0.5 ··· 0.9505 0.9599 ··· ··· ··· ··· ··· ··· .09 ··· 0.954 0.963 ··· Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z .00 .01 .02 .03 .04 0.5 ··· ··· ··· ··· ··· ··· ··· 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 ··· ··· ··· ··· ··· ··· ··· η(95) = 1.645, a linear interpolation of 1.64 and 1.65. ··· ··· ··· ··· ··· .09 ··· 0.954 0.963 ··· Normal Distribution e.g. Find the 95th percentile for the standard normal rv Z z .00 .01 .02 .03 .04 0.5 ··· ··· ··· ··· ··· ··· ··· 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 ··· ··· ··· ··· ··· ··· ··· η(95) = 1.645, a linear interpolation of 1.64 and 1.65. ··· ··· ··· ··· ··· Remark: If p does not appear in the table, we can either use the number closest to it, or use the linear interpolation of the closest two. .09 ··· 0.954 0.963 ··· Normal Distribution Normal Distribution In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation zα will denote the value on the z axis for which α of the area under the z curve lies to the right of zα . Normal Distribution In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation zα will denote the value on the z axis for which α of the area under the z curve lies to the right of zα . zα Normal Distribution Normal Distribution Remark: 1. zα is the 100(1 − α)th percentile of the standard normal distribution. Normal Distribution Remark: 1. zα is the 100(1 − α)th percentile of the standard normal distribution. 2. By symmetry the area under the standard normal curve to the left of −zα is also α. Normal Distribution Remark: 1. zα is the 100(1 − α)th percentile of the standard normal distribution. 2. By symmetry the area under the standard normal curve to the left of −zα is also α. 3. The zα s are usually referred to as z critical values. Percentile α (tail area) zα 90 0.1 1.28 95 0.05 1.645 97.5 0.025 1.96 ··· ··· ··· 99.95 0.0005 3.27 Normal Distribution Normal Distribution Proposition If X has a normal distribution with mean µ and stadard deviation σ, then X −µ Z= σ has a standard normal distribution. Thus a−µ b−µ ≤Z ≤ ) σ σ b−µ a−µ = Φ( ) − Φ( ) σ σ P(a ≤ X ≤ b) = P( P(X ≤ a) = Φ( a−µ ) σ P(X ≥ b) = 1 − Φ( b−µ ) σ Normal Distribution Normal Distribution Example (Problem 38): There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3cm and standard deviation 0.1cm. The second produces corks with diameters that have a normal distribution with mean 3.04cm and standard deviation 0.02cm. Acceptable corks have diameters between 2.9cm and 3.1cm. Which machine is more likely to produce an acceptable cork? Normal Distribution Example (Problem 38): There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3cm and standard deviation 0.1cm. The second produces corks with diameters that have a normal distribution with mean 3.04cm and standard deviation 0.02cm. Acceptable corks have diameters between 2.9cm and 3.1cm. Which machine is more likely to produce an acceptable cork? 2.9 − 3 3.1 − 3 ≤Z ≤ ) 0.1 0.1 = P(−1 ≤ Z ≤ 1) = 0.8413 − 0.1587 = 0.6826 2.9 − 3.04 3.1 − 3.04 P(2.9 ≤ X2 ≤ 3.1) = P( ≤Z ≤ ) 0.02 0.02 = P(−7 ≤ Z ≤ 3) = 0.9987 − 0 = 0.9987 P(2.9 ≤ X1 ≤ 3.1) = P( Normal Distribution Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value? Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value? µ − 1.5σ − µ µ + 1.5σ − µ ≤Z ≤ ) σ σ = P(−1.5 ≤ Z ≤ 1.5) P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P( = 0.9332 − 0.0668 = 0.8664 Normal Distribution Example (Problem 44): If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a)within 1.5 SDs of its mean value? (b)between 1 and 2 SDs from its mean value? µ − 1.5σ − µ µ + 1.5σ − µ ≤Z ≤ ) σ σ = P(−1.5 ≤ Z ≤ 1.5) P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P( = 0.9332 − 0.0668 = 0.8664 µ+σ−µ µ + 2σ − µ ≤Z ≤ ) σ σ = 2P(1 ≤ Z ≤ 2) 2 · P(µ + σ ≤ X1 ≤ µ + 2σ) = 2P( = 2(0.9772 − 0.8413) = 0.0.2718 Normal Distribution Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} · σ Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} · σ Example (Problem 39) The width of a line etched on an integrated circuit chip is normally distributed with mean 3.000 µm and standard deviation 0.140. What width value separates the widest 10% of all such lines from the other 90%? Normal Distribution Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} · σ Example (Problem 39) The width of a line etched on an integrated circuit chip is normally distributed with mean 3.000 µm and standard deviation 0.140. What width value separates the widest 10% of all such lines from the other 90%? ηN(3,0.1402 ) (90) = 3.0+0.140·ηN(0,1) (90) = 3.0+0.140·1.28 = 3.1792 Normal Distribution Normal Distribution Proposition Let X be a binomial rv based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, X has √ approximately a normal distribution with µ = np and σ = npq, where q = 1 − p. In particular, for x = a posible value of X , area under the normal curve P(X ≤ x) = B(x; n, p) ≈ to the left of x+0.5 x+0.5 − np = Φ( √ ) npq In practice, the approximation is adequate provided that both np ≥ 10 and nq ≥ 10, since there is then enough symmetry in the underlying binomial distribution. Normal Distribution Normal Distribution A graphical explanation for P(X ≤ x) = B(x; n, p) ≈ = Φ( area under the normal curve to the left of x+0.5 x+0.5 − np ) √ npq Normal Distribution A graphical explanation for P(X ≤ x) = B(x; n, p) ≈ = Φ( area under the normal curve to the left of x+0.5 x+0.5 − np ) √ npq Normal Distribution Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)? Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)? In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus np = 20 > 10 and nq = 180 > 10 Normal Distribution Example (Problem 54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is between 15 and 25 (inclusive)? In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus np = 20 > 10 and nq = 180 > 10 P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1) − Bin(14; 200, 0.1) 25 + 0.5 − 20 15 + 0.5 − 20 ≈ Φ( √ ) − Φ( √ ) 200 · 0.1 · 0.9 200 · 0.1 · 0.9 = Φ(0.3056) − Φ(−0.2500) = 0.6217 − 0.4013 = 0.2204