Topic 7 QBA241 2015

7. Continuous Probability Distribution

The probability distributions can be categorized into two types:

Probability Distributions

Discrete Continuous

Binomial Normal

Poisson

The topic of discrete probability distributions has been covered in the previous topic. This topic will
discuss the continuous probability distribution which is a normal distribution specifically.

A continuous random variable can assume an uncountable number of values.

7.1 Normal Distribution

The normal distribution is the most common statistical distributions, which has the following
properties.

Properties of a Normal Distribution

 It has a ‘bell shaped’ with a single peak

 The distribution is symmetrical (Mean = Median = Mode)
 The location is characterized by the mean (μ)
 The spread is characterized by the standard deviation, (σ)
 The right-side of the curve is the mirror image of the left-side of the curve (vice-versa)
 The random variable has an infinite theoretical range:  to +

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7.1.1 Shape of a Normal Distribution

The normal distribution is the most common statistical distributions, which has the following
properties.

Varying “Location” and “Spread”

By varying the parameters: μ and σ, we obtain different

normal distributions.

Same  but Different µ Same µ but Different 

7.2 Standard Normal Distribution

The simple case of a normal distribution is the standard normal distribution.

Any X random variable that is normally distributed (with any mean and standard deviation
combination) can be transformed into the standardized normal distribution (Z).

The standardized normal distribution has a mean of 0 and a standard deviation of 1.

Values above the mean have positive Z-values,

values below the mean have negative Z-values

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7.2.1 Transforming a Random Variable of X to a Standard Normal Distribution

Any random variables of X which is normally distributed can be converted to a standard normal
distribution known as the “Z values” or the “Z scores”.

Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and
dividing by its standard deviation.

Example 7.2.1

If X is distributed normally with mean of 100 and standard deviation of 50, the Z score for X =
200 is _____________

Solution
X ~ N ( = 100, σ = 50)

This means that X = 200 is two standard deviations (2 increments of 50 units) above the mean of
100.

To illustrate this, the curve is shown below.

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7.2.2 Probabilities of a Standard Normal Distribution

The total probability under the normal distribution curve is 1.00. Since the curve is symmetrical,
hence, the probability of the left-hand side and the right-hand side is equally 0.50.

In statistics, there is Empirical Rule which is also known as 68–95–99.7 rule. It is a shorthand used
to remember the percentage of values that lie within in a band around the mean in a normal
distribution with a width of one, two and three standard deviations, respectively; more accurately,
68.27%, 95.45% and 99.73% of the values lie within one, two and three standard deviations of the
mean, respectively.

Figure 1: Empirical Rule

FINDING PROBABILITY
Example 7.2.2 (a)

If X is distributed normally with mean of 100 and standard deviation of 50, what is the probability
when X is greater than 200?

Solution
X ~ N ( = 100, σ = 50)

( ) ( )

( )
( )
*Refer to the following Table A Standard Normal Probabilities.

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FINDING PROBABILITY
Example 7.2.2 (b)

Let X represent the time it takes (in seconds) to download an image file from the Internet. Suppose
X is normal with mean 8.0 and standard deviation of 5.0. Find the probability when the time it takes
to download an image file from the Internet is less than 8.6 seconds.

Solution
X ~ N ( = 8.0, σ = 5.0)

( ) ( ) ( )

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FINDING PROBABILITY
Example 7.2.2 (c)

Let X represent the time it takes (in seconds) to download an image file from the Internet. Suppose
X is normal with mean 8.0 and standard deviation of 5.0. Find the probability when the time it takes
to download an image file from the Internet is greater than 8.6 seconds.

Solution

X ~ N ( = 8.0, σ = 5.0)

( ) ( ) ( )

( )

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FINDING PROBABILITY BETWEEN TWO VALUES

Example 7.2.2 (d)

Let X represent the time it takes (in seconds) to download an image file from the Internet. Suppose
X is normal with mean 8.0 and standard deviation of 5.0. Find the probability when the time it takes
to download an image file from the Internet is between 8 to 8.6 seconds.

Solution
X ~ N ( = 8.0, σ = 5.0)

( ) ( )

( )
( ) ( )

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FINDING X VALUE BASED ON GIVEN PROBABILITY

Example 7.2.2 (e)

Let X represent the time it takes (in seconds) to download an image file from the Internet. Suppose
X is normal with mean 8.0 and standard deviation 5.0. Find X such that 20% of download times are
less than X.

Solution
X ~ N ( = 8.0, σ = 5.0)

So 20% of the download times from the distribution with mean 8.0 and standard deviation 5.0 are
less than 3.80 seconds.

P(Z < 0.84) = 0.2005  0.20 (the nearest to 20%)

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EXERCISES

1. Find the probability for the following questions. The first question is done for you. Sketch
the normal probability curve and shade the probability of the corresponding questions:
a) P(Z < 0.30) = 0.6179

0.6179

0 0.30 Z
b) P(Z < 1.96)

0 Z
c) P(Z > 1.28)

0 Z
d) P(Z > 1.65)

0 Z

Find the value of Z* based on the given probability. The first question is done for you.
e) P(Z < Z*) = 0.4013. Answer: The Z* is 0.25. So P(Z < 0.25) = 0.4013

0.4013

0.25 0 Z

f) P(Z < Z*) = 0.0228

0 Z
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g) P(Z > Z*) = 0.05

0 Z

h) P(Z < Z*) = 0.975

0 Z

2. Mariam earned a score of 940 on a national achievement test. The test score was normally
distributed with a mean of 850 and a standard deviation of 100. What proportion of students
had a higher score than Mariam?

3. Annual salaries for a medium-size company are approximately normally distributed with a
mean of AED50,000 and a standard deviation of AED20,000. What percentage of company
workers who work under AED40,000?

4. The test marks in a mathematics course are normally distributed with a mean of 70 and a
standard deviation of 10. The professor plans to award an A to the top 10% in the course.
Find out the least marks should you obtain in order to achieve an A.

5. Most graduate schools of business require applicants for admission to take the Graduate
Management Admission Council’s GMAT examination. Scores on the GMAT are roughly
normally distributed with a mean of 527 and a standard deviation of 112.
a) What is the probability of an individual scoring above 500 on the GMAT?
b) How high must an individual score on the GMAT in order to score in the highest 5%?

6. Mensa is an organization whose members possess IQs that are in the top 2% of the
population. It is known that IQs are normally distributed with a mean of 100 and a standard
deviation of 16. Find the minimum IQ needed to be Mensa member.

7. The annual rate of return on a mutual fund is normally distributed with a mean of 14% and
a standard deviation of 18%.
a) What is the probability that the fund returns more than 25% next year?
b) What is the probability that the fund loses money next year?

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