Standard Normal Curve Areas z

Transcription

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2 Table A.3 Standard Normal Curve Areas z

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4 Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61).

5 Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z

6 Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z P(Z 1.61) = ;

7 Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z P(Z 1.61) = ; P(Z > 1.12) = 1 P(Z 1.12) = = ;

8 Z N(0, 1), calculate (a)p(z 1.61); (b)p(z > 1.12); and (c)p( 1.12 < Z 1.61). z P(Z 1.61) = ; P(Z > 1.12) = 1 P(Z 1.12) = = ; P( 1.12 < Z 1.61) = P(Z 1.61) P(Z 1.12) = =

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10 Many tables for the normal distribution contain only the nonnegative part.

11 Many tables for the normal distribution contain only the nonnegative part. z What is P(Z < 1.63)?

12 Many tables for the normal distribution contain only the nonnegative part. z 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) = =

13 Many tables for the normal distribution contain only the nonnegative part. z 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) = =

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15 Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy

16 Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy Similarly, the (100p)th percentile of the standard normal rv Z is defined by η(p) 1 p = F (η(p)) = e y 2 /2 dy 2π

17 Recall: The (100p)th percentile of the distribution of a continuous rv X, η(p), is defined by p = F (η(p)) = η(p) f (y)dy Similarly, the (100p)th percentile of the standard normal rv Z is defined by η(p) 1 p = F (η(p)) = e y 2 /2 dy 2π We need to use the table for normal distribution to find (100p)th percentile.

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19 e.g. Find the 95th percentile for the standard normal rv Z

20 e.g. Find the 95th percentile for the standard normal rv Z z

21 e.g. Find the 95th percentile for the standard normal rv Z z η(95) = 1.645, a linear interpolation of 1.64 and 1.65.

22 e.g. Find the 95th percentile for the standard normal rv Z z η(95) = 1.645, a linear interpolation of 1.64 and 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.

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24 In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation z α will denote the value on the z axis for which α of the area under the z curve lies to the right of z α.

25 In statistical inference, the percentiles corresponding to right small tails are heavily used. Notation z α will denote the value on the z axis for which α of the area under the z curve lies to the right of z α. z α

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27 Remark: 1. z α is the 100(1 α)th percentile of the standard normal distribution.

28 Remark: 1. z α 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 z α is also α.

29 Remark: 1. z α 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 z α is also α. 3. The z α s are usually referred to as z critical values. Percentile α (tail area) z α

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31 Proposition If X has a normal distribution with mean µ and stadard deviation σ, then Z = X µ σ has a standard normal distribution. Thus P(a X b) = P( a µ Z b µ σ σ ) = Φ( b µ σ ) Φ(a µ σ ) P(X a) = Φ( a µ µ ) P(X b) = 1 Φ(b σ σ )

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33 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?

34 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? P(2.9 X 1 3.1) = P( Z ) = P( 1 Z 1) = = P(2.9 X 2 3.1) = P( Z ) = P( 7 Z 3) = =

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36 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?

37 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? P(µ 1.5σ X 1 µ + 1.5σ) = P( µ 1.5σ µ Z µ + 1.5σ µ ) σ σ = P( 1.5 Z 1.5) = =

38 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? P(µ 1.5σ X 1 µ + 1.5σ) = P( µ 1.5σ µ Z µ + 1.5σ µ ) σ σ = P( 1.5 Z 1.5) = = P(µ + σ X 1 µ + 2σ) = 2P( µ + σ µ Z µ + 2σ µ ) σ σ = 2P(1 Z 2) = 2( ) =

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40 Proposition {(100p)th percentile for N(µ, σ 2 )} = µ + {(100p)th percentile for N(0, 1)} σ

41 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 µm and standard deviation What width value separates the widest 10% of all such lines from the other 90%?

42 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 µm and standard deviation What width value separates the widest 10% of all such lines from the other 90%? η N(3, )(90) = η N(0,1) (90) = =

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44 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.

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46 A graphical explanation for ( ) 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

47 A graphical explanation for ( ) 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

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49 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)?

50 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

51 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) Φ( ) Φ( ) = Φ(0.3056) Φ( ) = =