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