MATH 166 Spring 2016
3.4
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3.4 The Normal Distribution
Definition: The random variable X has a standard normal distribution on the interval (−∞, ∞)
if the probability P (a ≤ X ≤ b) that X is between a and b is the area under the standard normal
curve given by
1
2
y = √ e−0.5x
2π
on the interval [a, b], where π ≈ 3.14159 and e ≈ 2.71828.
Remark: We will not use this formula in our calculations. Rather, we shall rely on tables or technology to find the area under this curve.
Characteristics of the Standard Normal Curve: The graph of the standard normal curve has
the following characteristics.
• It is bell shaped.
• It is symmetric about x = 0.
• It lies above the x-axis.
• It approaches but is never equal to 0 on both the positive and negative x-axis.
• The curve “bends downward” on the interval (−1, 1) and “bends upward” outside this interval.
• The area under the entire curve is exactly 1.
Note: The mean is µ = 0 and the standard deviation is σ = 1.
Convention: The random variable associated with the standard normal distribution is designated
by Z.
Note: Given any value b, a table on page 431 gives the area under the standard normal curve to the
left of b. We designated this area as A(b).
• If we are asked for P (Z ≤ b), this is just A(b). Thus
P (Z ≤ b) = A(b)
where we can find A(b) using the table/calculator.
• If we are asked for P (b ≤ Z), this is the area under the normal curve to the right of b, which is
P (b ≤ Z) = 1 − A(b)
• If we are asked for P (a ≤ Z ≤ b), this is the area under the normal curve from a to b, which is
P (a ≤ Z ≤ b) = A(b) − A(a)
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MATH 166 Spring 2016
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Technology Corner: To find the probability that a value of x is between lef t and right:
• The function is “normalcdf(left, right)”.
• The value ±1 × 1099 will function effectively as ±∞.
• To find P (Z ≤ 1.23) (note that the left endpoint is −1 × 1099 and the right endpoint is
1.23): Press 2ND— VARS — 2:normalcdf( and enter (-)— 1 —2ND— , —99— , — 1.23 —
) — ENTER .
• To find P (Z ≥ 1.23): Press 2ND— VARS — 2:normalcdf( and enter 1.23 — , — 1 —2ND—
, —99— ) — ENTER OR follow the steps as stated in previous part for P (Z ≤ 1.23) and
then enter “1-Ans”.
• To find P (−1 ≤ Z ≤ 1): Press 2ND— VARS — 2:normalcdf( and enter (-)— 1 — , — 1 —
) — ENTER .
Examples:
1. Choose a sketch of the area under the standard normal curve corresponding to P (0.56 < Z <
1.68) and Find the value of the probability of the standard normal variable Z corresponding to
P (0.56 < Z < 1.68).
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MATH 166 Spring 2016
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2. Find the indicated probability P (Z ≥ −1.59) given that Z is a random variable with a standard
normal distribution.
Technology Corner: The function to find c such that P (Z ≤ c) = 0.75 on a standard normal
probability distribution is “invNorm(p)”: Press 2ND— VARS — 3:invNorm( and enter 0.75.
Example: Let Z be the standard normal variable. Find the values of a that satisfy the given
probabilities.
1. P (Z < a) = 0.2818
2. P (Z > a) = 0.2992
3. P (−a < Z < a) = 0.4483
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Definition: The random variable X has a normal probability distribution with mean µ and standard
deviation σ on (−∞, ∞) if the probability P (a ≤ X ≤ b) that X is between a and b is the area under
the normal curve given by
1
2
y = √ e−0.5[(x−µ)/σ]
σ 2π
on the interval [a, b].
Remark: We will not use this formula in our calculations. Rather, we shall rely on tables or technology to find the area under this curve.
Characteristics of the Normal Curve: The graph of the normal curve has the following characteristics.
• It is bell shaped.
• It is symmetric about x = µ.
• It lies above the x-axis.
• It approaches but is never equal to 0 on both the positive and negative x-axis.
• The curve “bends downward” on the interval (µ − σ, µ + σ) and “bends upward” outside this
interval.
• The area under the entire curve is exactly 1.
Finding the Probability of a Normal Distribution: Let X be a random variable with mean µ
and standard deviation σ. Then
a−µ
b−µ
P (a ≤ X ≤ b) = P
≤Z≤
σ
σ
b−µ
a−µ
=A
−A
σ
σ
Technology Corner: To find the probability that a value of x is between lef t and right with a
mean of µ and a standard deviation of σ:
• The function is “normalcdf(left, right, µ, σ)”.
• To find P (X ≤ 95) with µ = 100 and σ = 4 (note that the left endpoint is −1 × 1099 and the
right endpoint is 95): Press 2ND— VARS — 2:normalcdf( and enter (-)— 1 —2ND— , —
99— , — 95 — , —100— , — 4 — ) — ENTER .
• To find P (X ≥ 95) with µ = 100 and σ = 4: Press 2ND— VARS — 2:normalcdf( and enter
95 — , — 1 —2ND— , —99— , —100— , — 4 — ) — ENTER .
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• To find P (0.99 ≤ X ≤ 1.01) with µ = 1 and σ = 0.004: Press 2ND— VARS — 2:normalcdf(
and enter 0.99— , —1.01— , — 1 — , — 0.004 — ) — ENTER .
• The function to find c such that P (X ≤ c) = 0.75 on a normal probability distribution
with a mean of 20 and standard deviation of 5 is “invNorm(p, µ, σ)”: Press 2ND— VARS —
3:invNorm( and enter 0.75,20,5,) .
Examples:
1. On the average, a student takes 91 words/minute midway through an advanced court reporting
course at the American Institute of Court Reporting. Assuming that the dictation speeds of
the students are normally distributed and that the standard deviation is 25 words/minute, find
the probability that a student randomly selected from the course can make dictation at the
following speeds.
(a) less than 41 words/minute
(b) more than 141 words/minute
(c) between 116 and 141 words/minute
2. A teacher wishes to “curve” a test whose grades were normally distributed with a mean of 66
and standard deviation of 12. The top 15% of the class will get an A, the next 25% of the class
will get a B, the next 30% of the class will get a C, the next 25% of the class will get a D and
the bottom 5% of the class will get an F. Find the cutoff for A and B.
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MATH 166 Spring 2016
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3. The distribution of heights of adult males is normally distributed with mean 68 inches and
standard deviation 2.7 inches. Answer the following.
(a) What minimum height is taller than 72% of all adult males?
(b) What 2 heights that are symmetric about the mean enclose the middle 82% of all heights?
4. Find the indicated quantities given that X is a normal random variable with a mean of 50 and
a standard deviation of 12.5.
(a) Find the value of b such that P (X ≤ b) = 0.3783.
(b) Find the value of c such that P (X ≥ c) = 0.9505.
(c) Find the values of A and B such that P (A ≤ X ≤ B) = 0.4581 if A and B are symmetric
about the mean.
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