Download Section 3.4 completed notes

c
YeongChyuan
Chung
Math 166, Fall 2016
3.4 The Normal Distribution
In this section, we will learn how to handle a particular continuous random
variable, in contrast with the discrete random variables that we dealt with
in the previous sections.
Definition: A probability density function, or PDF is a function that represents the probability for a continuous random variable.
Properties of PDF:
1. The function must be positive (or zero) everywhere and the total area
between the graph of the function and the x-axis is one.
2. The probability P (a ≤ X ≤ b) that a random variable is between two
values, say a and b, is equal to the area under the graph between the
boundaries a and b.
3. For a continuous random variable X and any value a, P (X = a) = 0
and thus P (X ≤ a) = P (X < a) and P (X ≥ a) = P (X > a).
Normal Distribution: The probability density function for a normal distribution is
(x−µ)2
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f (x) = √ e− 2σ2
σ 2π
where µ is the mean and σ is the standard deviation.
A normal distribution curve (with mean µ and standard deviation σ) is a
bell-shaped probability density function such that
1. The curve is symmetric about x = µ and attains its maximum there.
2. It approaches zero on both ends, but never equals zero.
3. The standard deviation determines how “steep” or “flat” the curve is.
The curve is flatter for larger σ.
Definition: A standard normal distribution is a normal distribution with
µ = 0 and σ = 1. We usually denote the random variable associated with
the standard normal distribution by Z.
The figure below is the curve of the probability density function for the
standard normal distribution:
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c
YeongChyuan
Chung
Math 166, Fall 2016
Remark: Generally, if X is normally distributed with mean µ and standard
deviation σ, then the variable (X − µ)/σ is standard normally distributed.
To find P (a ≤ Z ≤ b) (for a standard normal distribution):
1. Press 2ND and then VARS to get to the Distributions menu.
2. Select 2:normalcdf(. We will never use normalpdf.
3. Type a, comma, b. Close the parentheses and hit ENTER.
To enter ±∞, we usually enter 1E99 or -1E99. To enter “-”, press (-). To
enter “E”, press 2ND, and then the comma.
Example: Suppose Z is a random variable with a standard normal distribution. Find the following probabilities:
(a) P (−0.2 ≤ Z ≤ 0.4) = normalcdf (−0.2, 0.4) = .2347
(b) P (Z ≥ 0.5) = normalcdf (0.5, 1E99) = .3085
(c) P (Z < −0.5) = normalcdf (−1E99, −0.5) = .3085
To find P (a ≤ X ≤ b) where X is a normal random variable with
mean µ and standard deviation σ:
1. Press 2ND and then VARS to get to the Distributions menu.
2. Select 2:normalcdf( and hit ENTER.
3. Type a, comma, b, comma, µ, comma, σ. Close the parentheses and
hit ENTER.
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c
YeongChyuan
Chung
Math 166, Fall 2016
Example: Suppose X is a normally distributed random variable with µ =
10 and σ = 4. Find the following probabilities:
(a) P (3 ≤ X ≤ 8) = normalcdf (3, 8, 10, 4) = .2685
(b) P (X > 11) = normalcdf (11, 1E99, 10, 4) = .4013
Conversely, if we know P (X ≤ a) = p but a is unknown, then we can find
the corresponding a.
Suppose X is a normally distributed random variable with mean µ and
standard deviation σ. To find a such that P (X ≤ a) = p where p is
given:
1. Press 2ND and then VARS to get to the Distributions menu.
2. Select 3:invNorm and hit ENTER.
3. Type p, comma, µ, comma, σ. Close the parentheses and hit ENTER.
Note:
• The given probability must be of a form with inequality sign being
≤ or <. If P (X ≥ a) is given instead, first compute P (X ≤ a) =
1 − P (X ≥ a).
• It is useful to draw a normal curve to represent the given information.
Example: Find the values of a and b such that
(a) P (Z ≤ a) = 0.6
Ans: a = invN orm(0.6, 0, 1) = .2533.
(b) P (X ≥ a) = 0.3 where µ = 20 and σ = 5.
Ans: P (X < a) = 1−0.3 = 0.7 so a = invN orm(0.7, 20, 5) = 22.6220.
(c) P (a ≤ X ≤ b) = 0.4 where µ = 3 and σ = 2, and a and b are
symmetric about the mean µ.
Ans: P (X < a) = 1−0.4
= 0.3 by symmetry (draw a picture to see)
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so a = invN orm(0.3, 3, 2) = 1.9512,
P (X < b) = 0.3 + 0.4 = 0.7 so b = invN orm(0.7, 3, 2) = 4.0488 (or by
using symmetry, b = 3 + (3 − a) = 4.0488).
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c
YeongChyuan
Chung
Math 166, Fall 2016
Example: The amount of soda in a 16-ounce can is normally distributed
with a mean of 16 ounces and a standard deviation of 0.5 ounces. What
percentage of these bags have at most 15 ounces?
Ans: Let X be the amount of soda in a 16-ounce can. Then P (X ≤ 15) =
normalcdf (−1E99, 15, 16, 0.5) = .0228.
Example: The exam grades on a math exam are normally distributed with
a mean of 72 and a standard deviation of 15. What percentage of students
scored
(a) at least 90? (Round to two decimal places.)
Ans: Let X be the score of a student.
P (X ≥ 90) = normalcdf (90, 1E99, 72, 15) = .11507 so 11.51% of students scored at least 90.
(b) less than 60? (Round to two decimal places.)
Ans: P (X < 60) = normalcdf (−1E99, 60, 72, 15) = .21186 so 21.19%
of students scored less than 60.
(c) between 70 and 80? (Round to two decimal places.)
Ans: P (70 ≤ X ≤ 80) = normalcdf (70, 80, 72, 15) = .25613 so
25.61% of students scored between 70 and 80.
(d) What score would a student need to have made on the exam to be in
the top 10% of the class? (Round to two decimal places.)
Ans: P (X < a) = 0.9 (i.e., 90% of the class scored less than a) so
a = invN orm(0.9, 72, 15) = 91.22
(e) What score corresponds to the 75th percentile (meaning 75% of students scored less than this)? (Round to two decimal places.)
Ans: P (X < a) = 0.75 so a = invN orm(.75, 72, 15) = 82.12
Section 3.4 suggested homework: 3, 5, 7, 11, 13, 15, 17, 19, 23,
25, 27, 29, 31, 33
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