-Normal-Distribution-ppt.ppt-POWER PRESENTATION ON STATISTICS AND PROBABILITY
- 1. Normal Probability Distributions CARPIO, MUSK, BRIN - STAT. AND PROB. 2025
- 2. Properties of Normal Distributions A continuous random variable has an infinite number of possible values that can be represented by an interval on the number line. Hours spent studying in a day 0 6 3 9 15 12 18 24 21 The time spent studying can be any number between 0 and 24. The probability distribution of a continuous random variable is called a continuous probability distribution.
- 3. Properties of Normal Distributions The most important probability distribution in statistics is the normal distribution. A normal distribution is a continuous probability distribution for a random variable, x. The graph of a normal distribution is called the normal curve or Gaussian curve. Normal curve x
- 4. Properties of Normal Distributions Properties of a Normal Distribution 1. The mean, median, and mode are equal. 2. The normal curve is bell-shaped and symmetric about the mean. 3. The total area under the curve is equal to one. • The normal curve approaches, but never touches the x- axis as it extends farther and farther away from the mean. 1. Between μ σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points.
- 5. Properties of Normal Distributions μ 3σ μ + σ μ 2σ μ σ μ μ + 2σ μ + 3σ Inflection points Total area = 1 If x is a continuous random variable having a normal distribution with mean μ and standard deviation σ, you can graph a normal curve with the equation 2 2 ( ) 2 1 = 2 x μ σ y e σ π - - . = 2.178 = 3.14 e π x
- 6. Means and Standard Deviations A normal distribution can have any mean and any positive standard deviation. Mean: μ = 3.5 Standard deviation: σ 1.3 Mean: μ = 6 Standard deviation: σ 1.9 The mean gives the location of the line of symmetry. The standard deviation describes the spread of the data. Inflection points Inflection points 3 6 1 5 4 2 x 3 6 1 5 4 2 9 7 11 10 8 x
- 7. Means and Standard Deviations Example: 1. Which curve has the greater mean? 2. Which curve has the greater standard deviation? The line of symmetry of curve A occurs at x = 5. The line of symmetry of curve B occurs at x = 9. Curve B has the greater mean. Curve B is more spread out than curve A, so curve B has the greater standard deviation. 3 1 5 9 7 11 13 A B x
- 8. Interpreting Graphs Example: The heights of fully grown magnolia bushes are normally distributed. The curve represents the distribution. What is the mean height of a fully grown magnolia bush? Estimate the standard deviation. The heights of the magnolia bushes are normally distributed with a mean height of about 8 feet and a standard deviation of about 0.7 feet. μ = 8 REMEMBER! The inflection points are one standard deviation away from the mean. σ 0.7 6 8 7 9 10 Height (in feet) x
- 9. 3 1 2 1 0 2 3 z The Standard Normal Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Any value can be transformed into a z-score by using the formula The horizontal scale corresponds to z-scores. - - Value Mean = = . Standard deviation x μ z σ
- 10. The Standard Normal Distribution If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution. After the formula is used to transform an x-value into a z-score, the Standard Normal Table in the Z table is used to find the cumulative area under the curve. The area that falls in the interval under the nonstandard normal curve (the x- values) is the same as the area under the standard normal curve (within the corresponding z-boundaries). 3 1 2 1 0 2 3 z
- 12. The Standard Normal Table Properties of the Standard Normal Distribution 1. The cumulative area is close to 0 for z-scores close to z = 3.49. 2. The cumulative area increases as the z-scores increase. 3. The cumulative area for z = 0 is 0.5000. 4. The cumulative area is close to 1 for z-scores close to z = 3.49 z = 3.49 Area is close to 0. z = 0 Area is 0.5000. z = 3.49 Area is close to 1. z 3 1 2 1 0 2 3
- 13. Guidelines for Finding Areas Finding Areas Under the Standard Normal Curve 1. Sketch the standard normal curve and shade the appropriate area under the curve. 2. Find the area by following the directions for each case shown. a. To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table. 1. Use the table to find the area for the z-score. 2. The area to the left of z = 1.23 is 0.8907. 1.23 0 z
- 14. Guidelines for Finding Areas Finding Areas Under the Standard Normal Curve b. To find the area to the right of z, use the Standard Normal Table to find the area that corresponds to z. Then subtract the area from 1. 3. Subtract to find the area to the right of z = 1.23: 1 0.8907 = 0.1093. 1. Use the table to find the area for the z-score. 2. The area to the left of z = 1.23 is 0.8907. 1.23 0 z
- 15. Finding Areas Under the Standard Normal Curve c. To find the area between two z-scores, find the area corresponding to each z-score in the Standard Normal Table. Then subtract the smaller area from the larger area. Guidelines for Finding Areas 4. Subtract to find the area of the region between the two z-scores: 0.8907 0.2266 = 0.6641. 1. Use the table to find the area for the z-score. 3. The area to the left of z = 0.75 is 0.2266. 2. The area to the left of z = 1.23 is 0.8907. 1.23 0 z 0.75
- 16. Guidelines for Finding Areas Example: Find the area under the standard normal curve to the left of z = 2.33. From the Standard Normal Table, the area is equal to 0.0099. Always draw the curve! 2.33 0 z
- 17. Guidelines for Finding Areas Example: Find the area under the standard normal curve to the right of z = 0.94. From the Standard Normal Table, the area is equal to 0.1736. Always draw the curve! 0.8264 1 0.8264 = 0.1736 0.94 0 z
- 18. Guidelines for Finding Areas Example: Find the area under the standard normal curve between z = 1.98 and z = 1.07. From the Standard Normal Table, the area is equal to 0.8338. Always draw the curve! 0.8577 0.0239 = 0.8338 0.8577 0.0239 1.07 0 z 1.98
- 20. Probability and Normal Distributions If a random variable, x, is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval. P(x < 15) μ = 10 σ = 5 15 μ =10 x
- 21. Probability and Normal Distributions Same area P(x < 15) = P(z < 1) = Shaded area under the curve = 0.8413 15 μ =10 P(x < 15) μ = 10 σ = 5 Normal Distribution x 1 μ =0 μ = 0 σ = 1 Standard Normal Distribution z P(z < 1)
- 22. Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90. Probability and Normal Distributions P(x < 90) = P(z < 1.5) = 0.9332 - 90-78 = 8 x μ z σ = 1.5 The probability that a student receives a test score less than 90 is 0.9332. μ =0 z ? 1.5 90 μ =78 P(x < 90) μ = 78 σ = 8 x
- 23. Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85. Probability and Normal Distributions P(x > 85) = P(z > 0.88) = 1 P(z < 0.88) = 1 0.8106 = 0.1894 85-78 = = 8 x - μ z σ = 0.875 0.88 The probability that a student receives a test score greater than 85 is 0.1894. μ =0 z ? 0.88 85 μ =78 P(x > 85) μ = 78 σ = 8 x
- 24. Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80. Probability and Normal Distributions P(60 < x < 80) = P(2.25 < z < 0.25) = P(z < 0.25) P(z < 2.25) - - 1 60 78 = = 8 x μ z σ - = 2.25 The probability that a student receives a test score between 60 and 80 is 0.5865. 2 - - 80 78 = 8 x μ z σ = 0.25 μ =0 z ? ? 0.25 2.25 = 0.5987 0.0122 = 0.5865 60 80 μ =78 P(60 < x < 80) μ = 78 σ = 8 x