Density Curves & Normal Distribution Some Call it a Bell Curve - PowerPoint PPT Presentation

REVIEW PROBLEM Residents of upstate New York are accustomed to large amounts of snow with snowfalls often exceeding 6 inches in one day. In one city, such snowfalls were recorded for two seasons and are as follows (in inches): 8.6, 9.5, 14.1, 11.5, 7.0, 8.4, 9.0, 6.7, 21.5, 7.7, 6.8, 6.1, 8.5, 14.4, 6.1, 8.0, 9.2, 7.1 What are the mean and the population standard deviation for this data, to the nearest hundredth ? The mean = 9.46 inches. Standard deviation = 3.74

Density Curves & Normal Distribution Some Call it a “Bell Curve”

DENSITY CURVES Density curves give us a mathematical model for distributions. Density curves: are always on or above horizontal axis have an area of exactly 1 underneath it

Centers of Density Curves MEDIAN—equal-areas point MEAN—balance point The two are equal for a symmetric density curve. The mean is pulled towards the tail in a skewed curve.

In a skewed curve Median = Equal Areas Mean = Balance Point

NORMAL DISTRIBUTIONS We use (mu) to represent the mean and (sigma) to represent the standard deviation of a normal distribution. “Empirical Rule” 68-95-99.7 Rule

Normal Distribution A normal distribution is a continuous, symmetrical, bell-shaped distribution of a variable.

Characteristics of Normal Distribution 1. A normal distribution curve is bell-shaped. 2. The mean, median, and mode are equal and are located at the center of the distribution. 3. The curve is unimodal (i.e., it has only one mode) 4. The curve is symmetric about the mean (its shape is the same on both sides of a vertical line passing through the center) 5. The curve is continuous. There are no gaps or holes. For each value of X, there is a corresponding value of Y.

Characteristics of Normal Distribution 6. The curve never touches the x axis. No matter how far in either direction the curve extends, it never meets the x axis – but it gets increasingly closer. 7. The total area under a normal distribution curve is equal to 1.00, or 100%.

The area under the part of a normal curve that lies within 1 standard deviation of the mean is approximately 0.68, or 68%; within 2 standard deviations, about 0.95, or 95%; and within 3 standard deviations, about 0.997, or 99.7%.

The Empirical Rule (a.k.a. the “68 -95- 99.7 Rule”) In a normal distribution, almost all data will lie within 3 standard deviations of the mean. o About 68% of all data lies within 1 standard deviation of the mean. o About 95% of all data lies within 2 standard deviations of the mean. o About 99.7% of all data lies within 3 standard deviations of the mean.

Why Do We Need It? The Empirical Rule is most often used in statistics for forecasting or predicting final outcomes. After a standard deviation is calculated, and before exact data can be collected, the Empirical Rule can be used to estimate impending data. The probability based on the Empirical Rule can be used if gathering appropriate data may be time consuming, or even impossible to obtain.

Tips for Using the Empirical Rule Before applying the Empirical Rule it is a good idea to identify the data being described and the value of the mean and standard deviation. Sketch a graph summarizing the information provided by the empirical rule and identify the percentages for each region of the graph ( +/- 1, +/- 2, +/- 3) Remember that data must be normally distributed for the Empirical Rule to apply.

The Empirical Rule for a Normal Distribution

STEPS FOR FINDING NORMAL PROPORTION 1. State the problem in terms of x, your variable of concern. 2. Standardize the variable. 3. Draw a picture & shade area of concern. 4. Use the table to determine proportion.

From Proportion to Real Value GOING BACKWARDS! 1. State the problem. 2. Use the table. 3. Draw & shade a picture. 4. Unstandardize to obtain your x.

The amount of mustard dispensed from a machine at The Hotdog Emporium is normally distributed with a mean of 0.9 ounce and a standard deviation of 0.1 ounce. If the machine is used 500 times, approximately how many times will it be expected to dispense 1 or more ounces of mustard. Choose: 5 16 80 100 The mean is 0.9 and the standard deviation is 0.1. If one standard deviation is added to the mean, the result is 1.0 ounce. Therefore, dispensing 1 or more ounces falls into the category above one standard deviation to the right of the mean. Using the Empirical Rule, 16% of data falls at or above 1 standard deviation. 16% x 500 = 80 times to dispense one or more ounces of mustard.

A machine is used to fill soda bottles. The amount of soda dispensed into each bottle varies slightly. Suppose the amount of soda dispensed into the bottles is normally distributed. If at least 99% of the bottles must have between 585 and 595 milliliters of soda, find the greatest standard deviation, to the nearest hundredth, that can be allowed. The 99% implies a distribution within 3 standard deviations of the mean. The difference from 585 milliliters to 595 milliliters is 10 milliliters. Symmetrically divided, there are 5 milliliters used to create 3 standard deviations on one side of the mean. Dividing 5 by 3, we get the standard deviation to be 1.67 milliliters, to the nearest hundredth .

Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days. To the nearest percent, what percent of batteries have lifetimes less than 439 days? Subtracting, we see that 1 s.d. below is 439 days, an exact match to our question and an indication that the Empirical Rule can be used to find the answer. The question is asking what percent of a distribution is beyond one standard deviation to the left of the mean. Answer: 16%

A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal curve . If the mean shoe size is 11 and the standard deviation is 1.5, find: a. the probability that a man's shoe size is greater than or equal to 11. b. the probability that a man's shoe size is greater than or equal to 14. a. 50% In a normal distribution, the mean divides the data into two equal areas. Since 11 is the mean, 50% of the data is above 11 and 50% is below 11. b. 14 is exactly two standard deviations above the mean. Using the Empirical Rule we see that 2.5% will fall above two standard deviations. Probability is 0.025.

Five hundred values are normally distributed with a mean of 125 and a standard deviation of 10. a. What percent of the values lies in the interval 115 - 135, to the nearest percent ? b. What interval about the mean includes 95% of the data? a. What percent of the values is in the interval 115 - 135? mean + one standard deviation = 135 mean - one standard deviation = 115 Percent within one standard deviation of the mean = 68% b. 2 standard deviations about the mean for a total interval size of 40, with the mean in the center. mean + 2 standard deviations = 145 mean - 2 standard deviations = 105 Interval: [105,145]

Example: Estimating with the Empirical Rule You have purchased fluorescent light bulbs for your home. The average bulb life is 500 hours with a standard deviation of 24. The data is normally distributed. One of your bulbs burns out at 450 hours. Would you send the bulb back for a refund?

Problem: You have purchased fluorescent light bulbs for your home. The average bulb life is 500 hours with a standard deviation of 24. The data is normally distributed. One of your bulbs burns out at 450 hours. Would you send the bulb back for a refund? Solution: According to the Empirical Rule: o 68% of the light bulbs should last between 500 ± 24 or between 476 to 524 hours. o 95% of the light bulbs should last between 500 ± 2(24) or between 452 to 548 hours. o 99.7% of the light bulbs should last between 500 ± 3(24) or between 428 to 572 hours. If the light bulb only lasted 450 hours, I would consider it a defective bulb. Less than 2.5% should last less than 452 hours.

Your Turn The scores for all high school seniors taking the verbal section of the Scholastic Aptitude Test (SAT) in a particular year had a mean of 490 and a standard deviation of 100. The distribution of SAT scores is normal. 1. What percentage of seniors scored between 390 and 590 on this SAT test? 2. One student scored 795 on this test. How did this student do compared to the rest of the scores? 3. A rather exclusive university only admits students who were among the highest 16% of the scores on this test. What score would a student need on this test to be qualified for admittance to this university?

1. What percentage of seniors scored between 390 and 590 on this SAT test? The mean = 490 and the standard deviation = 100. So, we can draw the Normal Curve to depict the data. Calculate the difference between the mean and the test scores: |390-490| = 100  1 s.d. |590-490| = 100  1 s.d. So, about 68% of seniors would score between 390 and 590 on the SAT test.