Elements of Statistics II

Module 2:  Normal Distribution and Central Limit Theorem

Normal Distribution

The Normal distribution is a continuous probability distribution with a bell-shaped curve and its tails stretch infinitely in both directions. The distribution is symmetric on both sides from its mean (μ), its maximum frequency (height of the distribution or curve) is again at μ and its point of inflection is at its standard deviation (σ). A normal distribution is described by its mean (μ) and standard deviation (σ), and these two indices are the parameters of normal distribution, and they define the normal distribution completely. As the normal distribution is bell-shaped at its mean, the mean, median, and mode of a normal distribution coincide, that is, all have the same numerical value.

The normal distribution is described by the equation:

�(�)=1�2��−(�−�)22�2

-∞ < x < ∞

The area under the curve is the probability of the distribution.

Probability of 1 is the entire area under the curve.

�±�=68.26%

�±2�=95.44%

�±3�=99.73%

Normal Distribution Curve

 (Triola, 2018)

The Standard Normal Distribution

The standard normal distribution is a normal distribution with the parameters of μ = 0 and σ= 1. The total area under its density curve is equal to 1 (as shown in figure below).

Standard Normal Distribution

 (Triola, 2018)

 �∼�(�=0;�=1)

�=�−��

Example:

A bone mineral density test can be helpful in identifying the presence or likelihood of osteoporosis, a disease-causing bones to become more fragile and more likely to break. The result of a bone density test is commonly measured as a z score. The population of z scores is normally distributed with a mean of 0 and a standard deviation of 1, so these test results meet the requirements of a standard normal distribution, and the graph of the bone density test scores is as shown in Figure 6-5.

A randomly selected adult undergoes a bone density test. Find the probability that this person has a bone density test score less than 1.27.

A. Probability that the bone density test score is less than 1.27

B. Shaded area shown in Figure 6-5 So we need to find the area in Figure 6-5 below z = 1.27. If using technology, see the Tech Center instructions included at the end of this section. If using Table A-2, begin with the z score of 1.27 by locating 1.2 in the left column; next find the value in the adjoining row of probabilities that is directly below 0.07, as shown in the accompanying excerpt. Table A-2 shows that there is an area of 0.8980 corresponding to z = 1.27. We want the area below 1.27, and Table A-2 gives the cumulative area from the left, so the desired area is 0.8980. Because of the correspondence between area and probability, we know that the probability of a z score below 1.27 is 0.8980.

The probability that a randomly selected person has a bone μ density test result below 1.27 is 0.8980, shown as the shaded region in Figure 6-5. Another way to interpret this result is to conclude that 89.80% of people have bone density levels below 1.27.

 (Triola, 2018)

Central Limit Theorem

For all samples of the same size n with n ≥ 30, the sampling distribution of x̄ (sample mean) can be approximated by a normal distribution with mean μ and standard deviation �/�.

x̄∼�(��=��;��=��/�)

�∼�(�=0;�=1)

�=x̄−��/�

Practical Rules for Real Applications involving a sample mean:

Population has a normal distribution or n ≥ 30

Mean of all values of x̄: �x̄

Standard deviation of all values of x̄: �x̄=��

Z score conversion of x̄     �=x̄−���

If the original population is not normally distributed and n < 30, the distribution of x̄ cannot be approximated well by a normal distribution, and the methods of this section do not apply. Use other methods, such as nonparametric methods.

Example:

A random sample of 49 students indicates these students had a mean grade of 73 on a final exam, with a standard deviation of. Find the probability that the sample mean is less than 75.

Since the problem does not indicate that the sample comes from a population normally distributed and the sample size n ≥ 30, then the Central Limit Theorem can be applied.

P(x̄ < 75) , standardizing the variable �(x̄−���<75−73849)

P(Z < 21.14) = P(Z < 1.75), using the normal distribution table,

 P(Z < 1.75) = 0.9599

References

Textbooks (Suggested)

Rajaretnam, T. (2016).  Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

Triola, M. F. (2018).  Elementary statistics (13th ed.). Pearson. ISBN-13: 978-0134462455  https://librarylogin-carolina.uagm.edu/login?url=https://search.ebscohost.com/login.aspx?direct=true&db=e000xww&AN=1214457&site=ehost-live&ebv=EB&ppid=pp_Cover

Module 2:  Normal Distribution and Central Limit Theorem

Normal Distribution

The Normal distribution is a continuous probability distribution with a bell-shaped curve and its tails stretch infinitely in both directions. The distribution is symmetric on both sides from its mean (μ), its maximum frequency (height of the distribution or curve) is again at μ and its point of inflection is at its standard deviation (σ). A normal distribution is described by its mean (μ) and standard deviation (σ), and these two indices are the parameters of normal distribution, and they define the normal distribution completely. As the normal distribution is bell-shaped at its mean, the mean, median, and mode of a normal distribution coincide, that is, all have the same numerical value.

The normal distribution is described by the equation:

�(�)=1�2��−(�−�)22�2

-∞ < x < ∞

The area under the curve is the probability of the distribution.

Probability of 1 is the entire area under the curve.

�±�=68.26%

�±2�=95.44%

�±3�=99.73%

Normal Distribution Curve

 (Triola, 2018)

The Standard Normal Distribution

The standard normal distribution is a normal distribution with the parameters of μ = 0 and σ= 1. The total area under its density curve is equal to 1 (as shown in figure below).

Standard Normal Distribution

 (Triola, 2018)

 �∼�(�=0;�=1)

�=�−��

Example:

A bone mineral density test can be helpful in identifying the presence or likelihood of osteoporosis, a disease-causing bones to become more fragile and more likely to break. The result of a bone density test is commonly measured as a z score. The population of z scores is normally distributed with a mean of 0 and a standard deviation of 1, so these test results meet the requirements of a standard normal distribution, and the graph of the bone density test scores is as shown in Figure 6-5.

A randomly selected adult undergoes a bone density test. Find the probability that this person has a bone density test score less than 1.27.

A. Probability that the bone density test score is less than 1.27

B. Shaded area shown in Figure 6-5 So we need to find the area in Figure 6-5 below z = 1.27. If using technology, see the Tech Center instructions included at the end of this section. If using Table A-2, begin with the z score of 1.27 by locating 1.2 in the left column; next find the value in the adjoining row of probabilities that is directly below 0.07, as shown in the accompanying excerpt. Table A-2 shows that there is an area of 0.8980 corresponding to z = 1.27. We want the area below 1.27, and Table A-2 gives the cumulative area from the left, so the desired area is 0.8980. Because of the correspondence between area and probability, we know that the probability of a z score below 1.27 is 0.8980.

The probability that a randomly selected person has a bone μ density test result below 1.27 is 0.8980, shown as the shaded region in Figure 6-5. Another way to interpret this result is to conclude that 89.80% of people have bone density levels below 1.27.

 (Triola, 2018)

Central Limit Theorem

For all samples of the same size n with n ≥ 30, the sampling distribution of x̄ (sample mean) can be approximated by a normal distribution with mean μ and standard deviation �/�.

x̄∼�(��=��;��=��/�)

�∼�(�=0;�=1)

�=x̄−��/�

Practical Rules for Real Applications involving a sample mean:

Population has a normal distribution or n ≥ 30

Mean of all values of x̄: �x̄

Standard deviation of all values of x̄: �x̄=��

Z score conversion of x̄     �=x̄−���

If the original population is not normally distributed and n < 30, the distribution of x̄ cannot be approximated well by a normal distribution, and the methods of this section do not apply. Use other methods, such as nonparametric methods.

Example:

A random sample of 49 students indicates these students had a mean grade of 73 on a final exam, with a standard deviation of. Find the probability that the sample mean is less than 75.

Since the problem does not indicate that the sample comes from a population normally distributed and the sample size n ≥ 30, then the Central Limit Theorem can be applied.

P(x̄ < 75) , standardizing the variable �(x̄−���<75−73849)

P(Z < 21.14) = P(Z < 1.75), using the normal distribution table,

 P(Z < 1.75) = 0.9599

References

Textbooks (Suggested)

Rajaretnam, T. (2016).  Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560

Triola, M. F. (2018).  Elementary statistics (13th ed.). Pearson. ISBN-13: 978-0134462455  https://librarylogin-carolina.uagm.edu/login?url=https://search.ebscohost.com/login.aspx?direct=true&db=e000xww&AN=1214457&site=ehost-live&ebv=EB&ppid=pp_Cover