Normal Distribution

Table A.3 Standard Normal Curve Areas

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · ·

Normal Distribution

Table A.3 Standard Normal Curve Areas

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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 .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and(c)P(−1.12 < Z ≤ 1.61).

z .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and(c)P(−1.12 < Z ≤ 1.61).

z .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and(c)P(−1.12 < Z ≤ 1.61).

z .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and(c)P(−1.12 < Z ≤ 1.61).

z .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and(c)P(−1.12 < Z ≤ 1.61).

z .00 .01 .02 .03 .04 · · · .09· · · · · · · · · · · · · · · · · · · · · · · ·-1.2 0.1151 0.1131 0.1112 0.1094 0.1075 · · · 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 · · · 0.1170· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 · · · 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

Many tables for the normal distribution contain only thenonnegative part.

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 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 thenonnegative part.

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 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 thenonnegative part.

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 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 thenonnegative part.

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 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 thenonnegative part.

z .00 .01 .02 .03 .04 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 · · · 0.95451.7 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

Recall: The (100p)th percentile of the distribution of a continuousrv 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 isdefined by

p = F (η(p)) =

∫ η(p)

−∞

1√2π

e−y2/2dy

We need to use the table for normal distribution to find (100p)thpercentile.

Normal Distribution

Recall: The (100p)th percentile of the distribution of a continuousrv 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 isdefined by

p = F (η(p)) =

∫ η(p)

−∞

1√2π

e−y2/2dy

We need to use the table for normal distribution to find (100p)thpercentile.

Normal Distribution

Recall: The (100p)th percentile of the distribution of a continuousrv 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 isdefined by

p = F (η(p)) =

∫ η(p)

−∞

1√2π

e−y2/2dy

We need to use the table for normal distribution to find (100p)thpercentile.

Normal Distribution

Recall: The (100p)th percentile of the distribution of a continuousrv 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 isdefined by

p = F (η(p)) =

∫ η(p)

−∞

1√2π

e−y2/2dy

We need to use the table for normal distribution to find (100p)thpercentile.

Normal Distribution

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

z .00 .01 .02 .03 .04 0.5 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · · · · ·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 thenumber closest to it, or use the linear interpolation of the closesttwo.

Normal Distribution

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

z .00 .01 .02 .03 .04 0.5 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · · · · ·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 thenumber closest to it, or use the linear interpolation of the closesttwo.

Normal Distribution

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

z .00 .01 .02 .03 .04 0.5 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · · · · ·

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 thenumber closest to it, or use the linear interpolation of the closesttwo.

Normal Distribution

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

z .00 .01 .02 .03 .04 0.5 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · · · · ·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 thenumber closest to it, or use the linear interpolation of the closesttwo.

Normal Distribution

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

z .00 .01 .02 .03 .04 0.5 · · · .09

· · · · · · · · · · · · · · · · · · · · · · · · · · ·1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 · · · 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 · · · 0.9633· · · · · · · · · · · · · · · · · · · · · · · · · · ·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 thenumber closest to it, or use the linear interpolation of the closesttwo.

Normal Distribution

In statistical inference, the percentiles corresponding to right smalltails are heavily used.

Notationzα will denote the value on the z axis for which α of the areaunder the z curve lies to the right of zα.

zα

Normal Distribution

In statistical inference, the percentiles corresponding to right smalltails are heavily used.

Notationzα will denote the value on the z axis for which α of the areaunder the z curve lies to the right of zα.

zα

Normal Distribution

In statistical inference, the percentiles corresponding to right smalltails are heavily used.

Notationzα will denote the value on the z axis for which α of the areaunder the z curve lies to the right of zα.

zα

Normal Distribution

Remark:1. zα is the 100(1− α)th percentile of the standard normaldistribution.2. By symmetry the area under the standard normal curve to theleft of −zα is also α.3. The zαs are usually referred to as z critical values.

Percentile 90 95 97.5 · · · 99.95α (tail area) 0.1 0.05 0.025 · · · 0.0005

zα 1.28 1.645 1.96 · · · 3.27

Normal Distribution

Remark:1. zα is the 100(1− α)th percentile of the standard normaldistribution.

2. By symmetry the area under the standard normal curve to theleft of −zα is also α.3. The zαs are usually referred to as z critical values.

Percentile 90 95 97.5 · · · 99.95α (tail area) 0.1 0.05 0.025 · · · 0.0005

zα 1.28 1.645 1.96 · · · 3.27

Normal Distribution

Remark:1. zα is the 100(1− α)th percentile of the standard normaldistribution.2. By symmetry the area under the standard normal curve to theleft of −zα is also α.

3. The zαs are usually referred to as z critical values.

Percentile 90 95 97.5 · · · 99.95α (tail area) 0.1 0.05 0.025 · · · 0.0005

zα 1.28 1.645 1.96 · · · 3.27

Normal Distribution

Remark:1. zα is the 100(1− α)th percentile of the standard normaldistribution.2. By symmetry the area under the standard normal curve to theleft of −zα is also α.3. The zαs are usually referred to as z critical values.

Percentile 90 95 97.5 · · · 99.95α (tail area) 0.1 0.05 0.025 · · · 0.0005

zα 1.28 1.645 1.96 · · · 3.27

Normal Distribution

Proposition

If X has a normal distribution with mean µ and stadard deviationσ, then

Z =X − µσ

has a standard normal distribution. Thus

P(a ≤ X ≤ b) = P(a− µσ≤ Z ≤ b − µ

σ)

= Φ(b − µσ

)− Φ(a− µσ

)

P(X ≤ a) = Φ(a− µσ

) P(X ≥ b) = 1− Φ(b − µσ

)

Normal Distribution

Proposition

If X has a normal distribution with mean µ and stadard deviationσ, then

Z =X − µσ

has a standard normal distribution. Thus

P(a ≤ X ≤ b) = P(a− µσ≤ Z ≤ b − µ

σ)

= Φ(b − µσ

)− Φ(a− µσ

)

P(X ≤ a) = Φ(a− µσ

) P(X ≥ b) = 1− Φ(b − µσ

)

Normal Distribution

Example (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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 ≤ X1 ≤ 3.1) = P(2.9− 3

0.1≤ Z ≤ 3.1− 3

0.1)

= P(−1 ≤ Z ≤ 1) = 0.8413− 0.1587 = 0.6826

P(2.9 ≤ X2 ≤ 3.1) = P(2.9− 3.04

0.02≤ Z ≤ 3.1− 3.04

0.02)

= P(−7 ≤ Z ≤ 3) = 0.9987− 0 = 0.9987

Normal DistributionExample (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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 ≤ X1 ≤ 3.1) = P(2.9− 3

0.1≤ Z ≤ 3.1− 3

0.1)

= P(−1 ≤ Z ≤ 1) = 0.8413− 0.1587 = 0.6826

P(2.9 ≤ X2 ≤ 3.1) = P(2.9− 3.04

0.02≤ Z ≤ 3.1− 3.04

0.02)

= P(−7 ≤ Z ≤ 3) = 0.9987− 0 = 0.9987

Normal DistributionExample (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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 ≤ X1 ≤ 3.1) = P(2.9− 3

0.1≤ Z ≤ 3.1− 3

0.1)

= P(−1 ≤ Z ≤ 1) = 0.8413− 0.1587 = 0.6826

P(2.9 ≤ X2 ≤ 3.1) = P(2.9− 3.04

0.02≤ Z ≤ 3.1− 3.04

0.02)

= P(−7 ≤ Z ≤ 3) = 0.9987− 0 = 0.9987

Normal Distribution

Example (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?

P(µ− 1.5σ ≤ X1 ≤ µ+ 1.5σ) = P(µ− 1.5σ − µ

σ≤ Z ≤ µ+ 1.5σ − µ

σ)

= P(−1.5 ≤ Z ≤ 1.5)

= 0.9332− 0.0668 = 0.8664

2 · P(µ+ σ ≤ X1 ≤ µ+ 2σ) = 2P(µ+ σ − µ

σ≤ Z ≤ µ+ 2σ − µ

σ)

= 2P(1 ≤ Z ≤ 2)

= 2(0.9772− 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?

P(µ− 1.5σ ≤ X1 ≤ µ+ 1.5σ) = P(µ− 1.5σ − µ

σ≤ Z ≤ µ+ 1.5σ − µ

σ)

= P(−1.5 ≤ Z ≤ 1.5)

= 0.9332− 0.0668 = 0.8664

2 · P(µ+ σ ≤ X1 ≤ µ+ 2σ) = 2P(µ+ σ − µ

σ≤ Z ≤ µ+ 2σ − µ

σ)

= 2P(1 ≤ Z ≤ 2)

= 2(0.9772− 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?

P(µ− 1.5σ ≤ X1 ≤ µ+ 1.5σ) = P(µ− 1.5σ − µ

σ≤ Z ≤ µ+ 1.5σ − µ

σ)

= P(−1.5 ≤ Z ≤ 1.5)

= 0.9332− 0.0668 = 0.8664

2 · P(µ+ σ ≤ X1 ≤ µ+ 2σ) = 2P(µ+ σ − µ

σ≤ Z ≤ µ+ 2σ − µ

σ)

= 2P(1 ≤ Z ≤ 2)

= 2(0.9772− 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?

P(µ− 1.5σ ≤ X1 ≤ µ+ 1.5σ) = P(µ− 1.5σ − µ

σ≤ Z ≤ µ+ 1.5σ − µ

σ)

= P(−1.5 ≤ Z ≤ 1.5)

= 0.9332− 0.0668 = 0.8664

2 · P(µ+ σ ≤ X1 ≤ µ+ 2σ) = 2P(µ+ σ − µ

σ≤ Z ≤ µ+ 2σ − µ

σ)

= 2P(1 ≤ Z ≤ 2)

= 2(0.9772− 0.8413) = 0.0.2718

Normal Distribution

Proposition

{(100p)th percentile for N(µ, σ2)} =µ + {(100p)th percentile for N(0, 1)} · σ

Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 µm and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe 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

Proposition

{(100p)th percentile for N(µ, σ2)} =µ + {(100p)th percentile for N(0, 1)} · σ

Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 µm and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe 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

Proposition

{(100p)th percentile for N(µ, σ2)} =µ + {(100p)th percentile for N(0, 1)} · σ

Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 µm and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe 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

Proposition

{(100p)th percentile for N(µ, σ2)} =µ + {(100p)th percentile for N(0, 1)} · σ

Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 µm and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe 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

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 hasapproximately a normal distribution with µ = np and σ =

√npq,

where q = 1− p. In particular, for x = a posible value of X ,

P(X ≤ x) = B(x ; n, p) ≈(

area under the normal curve

to the left of x+0.5

)= Φ(

x+0.5− np√

npq)

In practice, the approximation is adequate provided that bothnp ≥ 10 and nq ≥ 10, since there is then enough symmetry in theunderlying binomial 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 hasapproximately a normal distribution with µ = np and σ =

√npq,

where q = 1− p. In particular, for x = a posible value of X ,

P(X ≤ x) = B(x ; n, p) ≈(

area under the normal curve

to the left of x+0.5

)= Φ(

x+0.5− np√

npq)

In practice, the approximation is adequate provided that bothnp ≥ 10 and nq ≥ 10, since there is then enough symmetry in theunderlying binomial distribution.

Normal Distribution

A graphical explanation for

P(X ≤ x) = B(x ; n, p) ≈(

area under the normal curve

to the left of x+0.5

)= Φ(

x+0.5− np√

npq)

Normal Distribution

A graphical explanation for

P(X ≤ x) = B(x ; n, p) ≈(

area under the normal curve

to the left of x+0.5

)= Φ(

x+0.5− np√

npq)

Normal Distribution

A graphical explanation for

P(X ≤ x) = B(x ; n, p) ≈(

area under the normal curve

to the left of x+0.5

)= Φ(

x+0.5− np√

npq)

Normal Distribution

Example (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1− p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10

P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1)− Bin(14; 200, 0.1)

≈ Φ(25 + 0.5− 20√

200 · 0.1 · 0.9)− Φ(

15 + 0.5− 20√200 · 0.1 · 0.9

)

= Φ(0.3056)− Φ(−0.2500)

= 0.6217− 0.4013

= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?

In this problem n = 200, p = 0.1 and q = 1− p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10

P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1)− Bin(14; 200, 0.1)

≈ Φ(25 + 0.5− 20√

200 · 0.1 · 0.9)− Φ(

15 + 0.5− 20√200 · 0.1 · 0.9

)

= Φ(0.3056)− Φ(−0.2500)

= 0.6217− 0.4013

= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1− p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10

P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1)− Bin(14; 200, 0.1)

≈ Φ(25 + 0.5− 20√

200 · 0.1 · 0.9)− Φ(

15 + 0.5− 20√200 · 0.1 · 0.9

)

= Φ(0.3056)− Φ(−0.2500)

= 0.6217− 0.4013

= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1− p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10

P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1)− Bin(14; 200, 0.1)

≈ Φ(25 + 0.5− 20√

200 · 0.1 · 0.9)− Φ(

15 + 0.5− 20√200 · 0.1 · 0.9

)

= Φ(0.3056)− Φ(−0.2500)

= 0.6217− 0.4013

= 0.2204