Discrete and Continuous Random Variables

Discrete and Continuous Random Variables

Continuous random variable: A variable whose values are not restricted.

12.5 – The Normal Distribution

Discrete random variable: A random variable that can take on only certain fixed values.

The number of even values of a single die. The number of heads in three tosses of a fair coin.

The diameter of a growing tree. The height of third graders.

Definition and Properties of a Normal Curve12.5 – The Normal Distribution

A normal curve is a symmetric, bell-shaped curve. Any random continuous variable whose graph has this characteristic shape is said to have a normal distribution.On a normal curve the horizontal axis is labeled with the mean and the specific data values of the standard deviations. If the horizontal axis is labeled using the number of standard deviations from the mean, rather than the specific data values, then the curve the standard normal curve


2.81.4 – 1 1 – 2 2 – 1.4 – 2.8

Normal Curve Standard Normal Curve

Sample Statistics

5.5 0 or 5.5

Normal Curves

0

C has mean > 0, standard deviation > 1

S

B

A

C


S is standard, with mean = 0, standard deviation = 1A has mean < 0, standard deviation = 1B has mean = 0, standard deviation < 1

Properties of Normal Curves

Empirical Rule: the approximate percentage of all data lying within 1, 2, and 3 standard deviations of the mean.


The graph of a normal curve is bell-shaped and symmetric about a vertical line through its center.The mean, median, and mode of a normal curve are all equal and occur at the center of the distribution.

within 3 standard deviations.

68%

95%

within 1 standard deviation

within 2 standard deviations

99.7%

Empirical Rule12.5 – The Normal Distribution

68%

95%99.7%

Example: Applying the Empirical RuleA sociology class of 280 students takes an exam. The distribution of their scores can be treated as normal. Find the number of scores falling within 2 standard deviations of the mean. A total of 95% of all scores lie within 2 standard deviations of the mean.


(.95)(280) =266 scores

Normal Curve Areas

Area (under the normal curve along an interval).


In a normal curve and a standard normal curve, the total area under the curve is equal to 1.The area under the curve is presented as one of the following:

Percentage (of total items that lie in an interval),Probability (of a randomly chosen item lying in an interval),

A Table of Standard Normal Curve Areas

The table shows the area under the curve for all values in a normal distribution that lie between the mean and z standard deviations from the mean.


To answer questions that involve regions other than 1, 2, or 3 standard deviations, a Table of Standard Normal Curve Areas is necessary.

Because of the symmetry of the normal curve, the table can be used for values above the mean or below the mean.

The percentage of values within a certain range of z-scores, or the probability of a value occurring within that range are the more common uses of the table.

Example: Applying the Normal Curve Table

Use the table to find the percent of all scores that lie between the mean and 1.5 standard deviations above the mean.


x z = 1.5z = 1.50

Therefore, 43.32% of all values lie between the mean and 1.5 standard deviations above the mean.

Find 1.50 in the z column. The table entry is .4332

There is a .4332 probability that a randomly selected value will lie between the mean and 1.5 standard deviations above the mean.

or

Example: Applying the Normal Curve TableUse the table to find the percent of all scores that lie between the mean and 2.62 standard deviations below the mean.


xz = –2.62

z = – 2.62

Therefore, 49.56% of all values lie between the mean and 2.62 standard deviations below the mean.

Find 2.62 in the z column. The table entry is 0.4956

There is a 0.4956 probability that a randomly selected value will lie between the mean and 2.62 standard deviations below the mean.

or

Example: Applying the Normal Curve TableFind the percent of all scores that lie between the given z-scores.


xz = –1.7

z = – 1.7

Therefore, 95% of all values lie between – 1.7 and 2.55 standard deviations.

The table entry is 0.4554

z = 2.55

z = 2.55 The table entry is 0.4946

0.4554 + 0.4946 = 0.95

Example: Applying the Normal Curve TableFind the probability that a randomly selected value will lie between the given z-scores.


x z = 0.61

z = 0.61

There is a 0.2666 probability that a randomly selected value will lie between 0.61 and 2.63 standard deviations.


z = 2.63

z = 2.63 The table entry is 0.4957

0.4957 – 0.2291 = 0.2666

Example: Applying the Normal Curve TableFind the probability that a randomly selected value will lie above the given z-score.


xz = 2.14

There is a 0.0162 probability that a randomly selected value will lie 2.14 standard deviations.


z = 2.14

Half of the area under the curve is 0.5000

0.5000 – 0.4838 = 0.0162

Example: Applying the Normal Curve Table


x

There is a 0.0139 probability that a randomly selected bottle will contain more than 12.33 ounces.

z = 2.2

12.33

Half of the area under the curve is 0.50000.5000 – 0.4861 = 0.0139

The volumes of soda in bottles from a small company are distributed normally with a mean of 12 ounces and a standard deviation .15 ounces. If 1 bottle is randomly selected, what is the probability that it will have more than 12.33 ounces?


Example: Finding z-scores for Given AreasAssuming a normal distribution, find the z-score meeting the condition that 39% of the area is to the right of z.

11%

39%


50% of the area lies to the right of the mean.= 0.11

= 0.39 The areas from the Normal Curve Table are based on the area between the mean and the z-score.

area between the mean and the z-score = 0.50 – 0.39 = 0.11From the table, find the area of 0.1100 or the closest value and read the z-score.

z-score = 0.28

Example: Finding z-scores for Given Areas


50% of the area lies to the left of the mean.

The areas from the Normal Curve Table are based on the area between the mean and the z-score.

area between the mean and the z-score = 0.76 – 0.50 = 0.26From the table, find the area of 0.2600 or the closest value and read the z-score.

z-score = 0.71

Assuming a normal distribution, find the z-score meeting the condition that 76% of the area is to the left of z.

26% = 0.2650%

0.5000

Discrete and Continuous Random Variables

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