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Normal Probability Distributions

Overview

Normal Distribution

- If a continuous random variable has a

distribution with a graph that is symmetric and

bell-shaped, and it can be described by the

equation below, we say that it has a normal

distribution.

The Normal Distribution

- The curve is bell-shaped and symmetric.

The Standard Normal Distribution

Uniform Distribution

- A continuous random variable has a uniform

distribution if its values spread evenly over the

range of possibilities. The graph of a uniform

distribution results in a rectangular shape.

Uniform Distribution

- The uniform distribution is symmetric and

rectangular.

Requirements for a Probability Distribution

- where x assumes all possible

values. - for every individual value of

x.

Density Curve

- A density curve is a graph of a continuous

probability distribution. It must satisfy the

following properties - The total area under the curve must equal 1.
- Every point on the curve must have a vertical

height that is 0 or greater. (That is, the curve

cannot fall below the x-axis.)

Relationship Between Area Under the Curve and

Probability

- Because the total area under a density curve is

equal to 1, there is a correspondence between

area and probability.

Example

- Suppose that the continuous random variable X has

a uniform distribution over the interval from 0

to 5. Find the probability that a randomly

selected value of X is - More than 4.
- Less than 2.
- Between 1 and 4.

Standard Normal Distribution

- The standard normal distribution is a normal

probability distribution that has a mean of 0 and

a standard deviation of 1, and the total area

under its density curve is equal to 1.

Standard Normal Distribution

- The standard normal distribution

Probability

- A probability of falling in an interval is just

the area under the curve.

Probability

Probabilities and a Continuous Probability

Distribution

- For continuous numerical variables and any

particular numbers a and b,

Calculating Probabilities Given a z Score

- Table A-2 is designed only for the standard

normal distribution, which has a mean of 0 and a

standard deviation of 0. - Table A-2 is on two pages, with one page for

negative z scores and the other page for positive

z scores. - Each value in the body of the table is a

cumulative area from the left up to a vertical

boundary above the specific z score.

Calculating Probabilities Given a z Score

- When working with a graph, avoid confusion

between z scores and areas. - z score Distance along the horizontal scale of

the standard normal distribution refer to the

leftmost column and top row of Table A-2. - Area Region under the curve refer to the values

in the body of Table A-2. - The part of the z score denoting hundredths is

found across the top row of Table A-2.

Example

- Find the area under the standard normal

distribution to the left of 1.5. - Find the area under the standard normal

distribution to the right of -2. - Find the area under the standard normal

distribution between -2 and 1.5.

Example

- Let z denote a random variable that has a

standard normal distribution. Determine each of

the following probabilities

Calculating a z Score Given a Probability

- Draw a bell-shaped curve and identify the region

under the curve that corresponds to the given

probability. If that region is not a cumulative

region from the left, work instead with a known

region that is a cumulative region from the left. - Using the cumulative area from the left, locate

the closest probability in the body of Table A-2

and identify the corresponding z score.

Example

- Determine the z value that separates
- the smallest 10 of all the z values from the

others, - the largest 5 of all the z values from the

others.

Applications of Normal Distributions

Standardizing Scores

- If we convert values to scores using

,then

procedures with all normal distributions are the

same as those for the standard normal

distribution.

Converting Values in a Nonstandard Normal

Distribution to z Scores

- Sketch a normal curve, label the mean and the

specific z values, then shade the region

representing the desired probability. - For each relevant value x that is a boundary for

the shaded region, use the z Score formula to

convert that value to the equivalent z score. - Refer to Table A-2 and use the z scores to find

the area of the shaded region. This area is the

desired probability.

Example

- The serum cholesterol levels of 17-year-olds

follow a normal distribution with a mean of 176

mg/dLi and a standard deviation of 30 mg/dLi. If

a 17-year-old is selected at random, what is the

probability he/she has a serum cholesterol level - of 156 mg/dLi or less?
- of more than 216 mg/dLi?
- between 121 mg/dLi and 186 mg/dLi?

z Scores and Area

- Dont confuse z scores and areas.
- Choose the correct side of the graph.
- A z score must be negative whenever it is located

in the left half of the normal distribution. - Areas (or probabilities) are positive or zero

values, but they are never negative.

Finding Values From Known Areas

- Sketch a normal distribution curve, enter the

given probability or percentage in the

appropriate region of the graph, and identify the

x value(s) being sought. - Use Table A-2 to find the z score corresponding

to the cumulative left area bounded by x. Refer

to the body of Table A-2 to find the closest

area, then identify the corresponding z score.

Finding Values From Known Areas

- Using the formula

,enter the values for , ,

and the z score found in Step 2, then solve for

x. - Refer to the sketch of the curve to verify that

the solution makes sense in the context of the

graph and in the context of the problem.

Example (continued)

- The serum cholesterol levels of 17-year-olds

follow a normal distribution with a mean of 176

mg/dLi and a standard deviation of 30 mg/dLi.

Find - the 80th percentile.
- the 25th percentile.

Sampling Distributions and Estimators

Sampling Distribution of the Mean

- The sampling distribution of the mean is the

probability distribution of sample means, with

all samples having the same sample size n.

Example

- Suppose our population consists of the three

values 1, 2, and 5. - Calculate the mean, median, range, variance and

standard deviation for the population. - Find all possible samples of 2 values.
- Calculate the mean, median, range, variance and

standard deviation for each sample. - Calculate the mean of the sample means, sample

medians, sample ranges, sample variances, and

sample standard deviation. - Compare the results of d with the results of a.

Example (continued)

Sampling Variability

- The value of a statistic, such as the sample mean

, depends on the particular values included

in the sample, and it generally varies from

sample to sample. This variability of a statistic

is called sampling variability.

Sampling Distribution of the Proportion

- The sampling distribution of the proportion is

the probability distribution of sample

proportions, with all samples having the same

sample size n.

Properties of the Distribution of Sample

Proportions

- Sample proportions tend to target the value of

the population proportion. - Under certain conditions, the distribution of

sample proportions approximates a normal

distribution.

Biased and Unbiased Estimators

- A sample statistic is an unbiased estimator of a

population parameter if it targets the

population parameter. - A sample statistic is a biased estimator of a

population parameter if it does not target the

population parameter.

Estimators Good and Bad

- Statistics that target population parameters

Mean, Variance, Proportion - Statistics that do not target population

parameters Median, Range, Standard Deviation

The Central Limit Theorem

Example

- The serum cholesterol levels of 17-year-olds

follow a normal distribution with a mean of 176

mg/dLi and a standard deviation of 30 mg/dLi. If

a 17-year-old is selected at random, what is the

probability he/she has a serum cholesterol level

of 156 mg/dLi or less?

The Central Limit Theorem and the Sampling

Distribution of

- Given
- The random variable x has a distribution (which

may or may not be normal) with mean and

standard deviation . - Simple random samples all of the same size n are

selected from the population. (The samples are

selected so that all possible samples of size n

have the same chance of being selected.)

The Central Limit Theorem and the Sampling

Distribution of

- Conclusions
- The distribution of sample means will, as the

sample increases, approach a normal distribution. - The mean of all sample means is the population

mean . (That is, the normal distribution from

Conclusion 1 has mean .) - The standard deviation of all sample means is

. (That is, the normal distribution from

Conclusion 1 has standard deviation .)

The Central Limit Theorem and the Sampling

Distribution of

- Practical Rules Commonly Used
- If the original population is not itself normally

distributed, here is a common guideline For

samples of size n greater than 30, the

distribution of the sample means can be

approximated reasonably well by a normal

distribution. (There are exceptions, such as

populations with very non-normal distributions

requiring samples sizes much larger than 30, but

such exceptions are relatively rare.) The

approximation gets better as the sample size n

becomes larger. - If the original population is itself normally

distributed, then the sample means will be

normally distributed for any sample size n (not

just the values of n larger than 30).

Notation for the Sampling Distribution of

- If all possible random samples of size n are

selected from a population with mean and

standard deviation , the mean of the sample

means is denoted by , soAlso, the

standard deviation of the samples means is

denoted by , so is often called the

standard error of the mean.

Example

- The serum cholesterol levels of 17-year-olds

follow a normal distribution with a mean of 176

mg/dLi and a standard deviation of 30 mg/dLi. If

a random sample of ten 17-year-olds is selected,

what is the probability that the sample mean is

156 mg/dLi or less?

The Central Limit Theorem - Bottom Line

- As the sample size increases, the sampling

distribution of sample means approaches a normal

distribution.

Example

- If the mean and standard deviation of serum iron

values for healthy men are 120 and 15 micrograms

per 100 mL, respectively, what is the probability

that a random sample of 50 healthy men will yield

a sample mean between 115 and 125 micrograms per

100 mL?

Applying the Central Limit Theorem

- When working with an individual value from a

normally distributed population, use the methods

of Section 5.3. Use - When working with a mean for same sample (or

group), be sure to use the value for

the standard deviation of the sample means. Use

Interpreting Results

- Rare Event RuleIf, under a given assumption, the

probability of a particular observed event is

exceptionally small, we conclude that the

assumption is probably not correct.

Correction for a Finite Population

- When sampling with replacement and the sample

size n is greater than 5 of the finite

population size N (that is, ),

adjust the standard deviation of the sample means

by multiplying it by the finite population

correction factor

Normal as Approximation to the Binomial

Recall

- A binomial probability distribution results from

a procedure that meets all the following

requirements - The procedure has a fixed number of trials.
- The trials must be independent.
- Each trial must have all outcomes classified into

two categories. - The probabilities must remain constant for each

trial.

Binomial Distributions p 0.5 n 3, n 4, n

5, n 6

Binomial Distributions p 0.5 n 10, n 20,

n 30, n 40

Binomial Distributions p 0.3 n 10, n 20,

n 30, n 40

Normal Distribution as Approximation to Binomial

Distribution

- When working with a binomial distribution, if

and , then the binomial random

variable has a probability distribution that can

be approximated by a normal distribution with the

mean and standard deviation given as

Definition

- When we use the normal distribution (which is a

continuous probability distribution) as an

approximation to the binomial distribution (which

is discrete), a continuity correction is made to

a discrete whole number x in the binomial

distribution by representing the single x value

by the interval from to

(that is, by adding and subtracting 0.5).

Example

- In a certain population of mussels (Mytilus

edulis), 80 of the individuals are infected with

an intestinal parasite. A marine biologist plans

to examine 100 randomly chosen mussels from the

population. Find the probability that at least 85

of the mussels will be infected.

Assessing Normality

Definition

- A normal quantile plot (or normal probability

plot) is a graph of the points (x, y) where each

x value is from the original set of sample data,

and each y value is the corresponding z score

that is a quantile value expected from the

standard normal distribution.

Procedure for Determining Whether Data Have a

Normal Distribution

- Histogram Construct a histogram. Reject

normality if the histogram departs dramatically

from a bell shape. - Outliers Identify outliers. Reject normality if

there is more than one outlier present. - Normal quantile plot If the histogram is

basically symmetric and there is at most one

outlier, construct a normal quantile plot.

Examine the normal quantile plot using these

criteria - If the points do not lie close to a straight

line, or if the points exhibit some systematic

pattern that is not a straight-line pattern, then

the data appear to come from a population that

does not have a normal distribution. - If the pattern of the points is reasonably close

to a straight line, then the data appear to come

from a population that has a normal distribution.

Example

- Recall our study of bears, the data for the

lengths of bears is given in Data Set 6 of

Appendix B. Determine whether the requirement of

a normal distribution is satisfied. Assume that

this requirement is loose in the sense that the

population distribution need not be exactly

normal, but it must be a distribution that is

basically symmetric with only one mode.

Example (continued)

Example (continued)

Example (continued)

- Using the weights of bears (given in Data Set 6

of Appendix B), determine whether the requirement

of a normal distribution is satisfied. Assume

that this requirement is loose in the sense that

the population distribution need not be exactly

normal, but it must be a distribution that is

basically symmetric with only one mode.

Example (continued)

Example (continued)

Data Transformations

- For data sets where the distribution is not

normal, we can transform the data so that the

modified values have a normal distribution.

Common transformations include