CONTINUOUS PROBABILITY
DISTRIBUTION
FS0102 INTRODUCTION TO STATISTICS
LECTURE 7
GOALS
1.
Understand the difference between discrete and continuous distributions.
2.
Compute the mean and the standard deviation for a uniform distribution.
3.
Compute probabilities by using the uniform distribution.
4.
List the characteristics of the normal probability distribution.
5.
Define and calculate z values.
(Reference: Chapter 7 of Textbook)
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Discrete VS Continuous Distributions
Discrete Probability Distribution
Continuous Probability Distribution
The Normal Distribution - Graphically
1.
It is bell-shaped and has a single peak at the center of the distribution.
2.
Mean = Median = Mode
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The Normal Distribution - Graphically
4.
It is symmetrical about the mean.
5.
under the normal curve is to the right of this center point and the other half to the left
of it.
6.
The total area under the curve is 1.00; half the area 0.5
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The Normal Distribution - Graphically
6.
It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. To
put it another way, the tails of the curve extend indefinitely in both directions.
7.
The location of a normal distribution is determined by the mean(), the dispersion or spread
of the distribution is determined by the standard deviation (σ) .
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The Normal Distribution - Families
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You Try: Practice 7A
Match the following to the items
labelled A-C:
• 30%
• 50%
• 100%
• Mean
• Variance
• Standard Deviation
Total area
under graph
= A%
Area to the right
of the middle
= B%
C
Z
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You Try: Practice 7A
Match the following to the
correct coloured graph:
• Lowest mean
• Highest standard deviation
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The Empirical Rule
• About 68 percent of the
area under the normal
curve is within one
standard deviation of
the mean.
• About 95 percent is
within two standard
deviations of the mean.
• Practically all (99.7%) is
within three standard
deviations of the mean.
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The Standard Normal
Probability Distribution
• The standard normal distribution
is a normal distribution with a
mean of 0 and a standard
deviation of 1.
• It is also called the 𝑧 distribution.
• A 𝑧-value is the signed distance
between a selected value,
designated 𝑋, and the
population mean , divided by
the population standard
deviation, σ.
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The Standard Normal
Probability Distribution
• The 𝑧-value is the ratio of difference between a selected
value and population mean and σ. i.e. 𝑧-value is the
difference of a selected value and mean express as the
number of times of σ.
• E.g. z = 1.5 means the difference is 1.5 times of σ
• The formula is:
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Standardize Normal Distribution
• Suppose we have a normal distribution. We
𝑋−𝜇
apply 𝑍 = 𝜎
• This normal distribution is then transformed
from N(μ , σ2). to N(0, 1):
• When 𝑥 = 𝜇 then 𝑧 =
𝜇–𝜇
𝜎
When 𝑥 = 𝜇 + 𝜎 then 𝑧 =
= 0.
𝜇+𝜎 −𝜇
𝜎
= 1.
This process is called standardization
of normal distribution.
This can be seen in the diagram:
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The Normal Distribution – Example
The weekly incomes of shift
foremen in the glass
industry follow the normal
probability distribution with
a mean of $1,000 and a
standard deviation of $100.
What is the z value for the
income, let’s call it X, of a
foreman who earns $1,100
per week? For a foreman
who earns $900 per week?
For X = 1100 :
X −
1100 − 1000
= + 1.00

100
For X = 900 :
Z=
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Z=
X −

=
=
900 − 1000
= − 1.00
100
Normal Distribution – Finding Probabilities
In an earlier example we
reported that the mean
weekly income of a shift
foreman in the glass
industry is normally
distributed with a mean of
$1,000 and a standard
deviation of $100.
Z
What is the probability of
selecting a foreman whose
weekly income is between
$1,000 and $1,100?
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X
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Normal Distribution – Finding Probabilities
X
Z
X
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Normal Distribution – Finding Probabilities (Example 2)
Refer to the information regarding
the weekly income of shift foremen
in the glass industry. The distribution
of weekly incomes follows the
normal probability distribution with a
mean of $1,000 and a standard
deviation of $100.
What is the probability of selecting a
shift foreman in the glass industry
whose income is:
Between $790 and $1,000?
Less than $790?
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Normal Distribution – Finding Probabilities (Example 2)
Refer to the information regarding
the weekly income of shift foremen
in the glass industry. The distribution
of weekly incomes follows the
normal probability distribution with a
mean of $1,000 and a standard
deviation of $100.
What is the probability of selecting a
shift foreman in the glass industry
whose income is:
Between $790 and $1,000?
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Z
X
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Normal Distribution – Finding Probabilities (Example 3)
Refer to the information regarding
the weekly income of shift foremen
in the glass industry. The
distribution of weekly incomes
follows the normal probability
distribution with a mean of $1,000
and a standard deviation of $100.
What is the probability of selecting
a shift foreman in the glass
industry whose income is:
Less than $790?
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You Try! (Offline Practice)
Refer to the information
regarding the weekly income of
shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability distribution
with a mean of $1,000 and a
standard deviation of $100.
What is the probability of
selecting a shift foreman in the
glass industry whose income is:
Between $840 and $1,200?
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Z
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Normal Distribution – Finding Probabilities (Example 4)
Refer to the information regarding the weekly
income of shift foremen in the glass industry. The
distribution of weekly incomes follows the normal
probability distribution with a mean of $1,000 and a
standard deviation of $100.
What is the probability of selecting a shift foreman
in the glass industry whose income is:
Between $840 and $1,200?
𝑃 𝑧 = −1.6 = 0.4452,
𝑃 𝑧 = 2 = 0.4772
𝑃 840 ≤ 𝑥 ≤ 1200 = 0.4452 + 0.4772
= 0.9224
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Z
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You Try! (Offline Practice)
Refer to the information regarding the
weekly income of shift foremen in the
glass industry. The distribution of weekly
incomes follows the normal probability
distribution with a mean of $1,000 and a
standard deviation of $100.
What is the probability of selecting a shift
foreman in the glass industry whose
income is:
Between $1,150 and $1,250
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Normal Distribution – Finding Probabilities (Example 5)
Refer to the information regarding the
weekly income of shift foremen in the
glass industry. The distribution of weekly
incomes follows the normal probability
distribution with a mean of $1,000 and a
standard deviation of $100.
What is the probability of selecting a shift
foreman in the glass industry whose
income is:
Z
Between $1,150 and $1,250
𝑃 𝑧 = 1.5 = 0.4332,
𝑃 1150 ≤ 𝑥 ≤ 1250
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𝑃 𝑧 = 2.5 = 0.4938
= 0.4938 − 0.4332 = 0.0606
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XZP Problems – A Summary
1. Draw diagram
a)
b)
c)
Mean in the m____
Place given 𝑥 value (what is 𝑥?)
Shade the area corresponding to
the required probability.
2. Find z value: 𝑧 =
a)
𝑧 is the ________________
3. Use 𝑧 to find the p_______ from
the n______________
4. Check again the required a____
5. Add or subtract accordingly for
the required probability.
or
You Try – Practice 7B
Q
P
The masses of chocolates sold at Willy’s Choc
Shop follows a normal distribution with 𝜇 =
320g and 𝜎 = 15.7g.
A
R
B
S
C
1.
What is the probability that Charlie will buy a
packet of chocolate weighing less than 350g?
Sketch the normal distribution curve,
indicating on the diagram the location of
350g and shade the required area.
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1.
2.
3.
2.
The mean is located at _____.
The value of the mean is _____.
I will put "350g" at _____.
Referring to the diagram above,
the area that corresponds to the
probability that "Charlie will buy a
packet of chocolate weighing less
than 350g" is __________.
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You Try – Practice 7B
Q
P
The masses of chocolates sold at Willy’s Choc
Shop follows a normal distribution with 𝜇 =
320g and 𝜎 = 15.7g.
What is the probability that Charlie will buy a
packet of chocolate weighing less than 350g?
Calculate the value of z, find its
corresponding probability value and hence
find P(weight < 350g).
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A
R
B
S
C
Question 3
a) Write your answer for z correct
to 2 decimal places.
z = _________
b) Write your answer for P(z)
correct to 4 decimal places.
From table, P(z) = _________
c)
P(Charlie will buy a packet of
chocolate weighing less than
350g) = _________
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Using Z in Finding X Given Area - Example
Layton Tire and Rubber Company
wishes to set a minimum mileage
guarantee on its new MX100 tire.
Tests reveal the mean mileage is
67,900 with a standard deviation of
2,050 miles and that the
distribution of miles follows the
normal probability distribution. It
wants to set the minimum
guaranteed mileage so that no
more than 4 percent of the tires
will have to be replaced. What
minimum guaranteed mileage
should Layton announce?
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Mean mileage is 67,900, standard deviation of 2,050
Lowest 4%
Using Z in Finding X Given Area - Example
• 4% mileage  on the left
• However, the z value corresponds to
P = 0.5 − 0.04 = 0.46
• From table,
𝑃 = 0.46 → 𝑍 = −1.75
• Negative since z is to the left of the mean
• When 𝑍 = −1.75,
−1.75 =
𝑋 − 67900
2050
𝑋 = 67900 − 1.75 × 2050
𝑋 = 64312
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You Try – Practice 7C
If the mean height of JCU students is
1.62 m with a standard deviation of
0.095 m,
a) which area corresponds to the
tallest 4.75% of the student
population?
b) Let the top 4.75% of the student
population be of at least height h.
What is the probability that a
randomly chosen student has a height
between mean and x, P(μ < x < h)
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You Try – Practice 7C
If the mean height of JCU students is
1.62 m with a standard deviation of
0.095 m,
c) State the z value that corresponds
to the least height of the top 4.75% of
the student population.
d) Calculate the least height of the
top 4.75% of the student population.
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