Normal Distribution S-ID.4 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.

Measures of Central Tendency Mean (𝝁): “average” add data values and divide by the number of values Median: list data values in order from least to greatest and find the middle Mode: most frequently occurring value Bimodal: when a set of data has two modes Outlier: a value that is substantially different from the rest of the data in a set

Example Find the mean, median and mode for the sets of data. Identify any outliers. 75, 68, 43, 120, 65, 180, 95, 225, 140 3.4, 4.5, 2.3, 5.9, 9.8, 3.3, 2.1, 3.0, 2.9

Standard Deviation and Variance Measures showing how much data values deviate from the mean (spread) σ (sigma): standard deviation 𝜎 2 (sigma squared): variance In calculator: type values into L1 (STAT – EDIT) STAT – CALC - #1 (gives all stats about that data)

Normal Distributions Contains data that varies randomly from the mean EX: test scores, weight, height, etc. Normal curve: the graph of a normal distribution Mean in the middle “bell curve” shape 3 standard deviations above and below the mean

Skew Outliers cause a normal curve to be right skewed or left skewed If model is skewed, measures of central tendency are NOT the same

Empirical Rule “68-95-99.7” 68% of data between ± 1 standard deviations of mean 95% of data between ± 2 standard deviations of mean 99.7% of data between ± 3 standard deviations of mean

Example The number of hours that students studied for final exams was normally distributed. Of the 200 students, the mean number of hours they studied was 12 hours. The standard deviation was 3 hours. Draw a normal curve to represent the distribution. What percentage of students studied 3 hours or less? Of the students surveyed, how many studied less than 9 hours?

Example The heights of girls in a school choir are distributed normally, with a mean of 64 and a standard deviation of 1.75. If 83 girls are between 60.5 in. and 67.5 in. tall, how many girls are in the choir?

Z-score The number of standard deviations a data value is from the mean Formula: 𝑧= 𝑥−𝜇 𝜎 EX: A teacher gave a test that had a class average of 85 with a standard deviation of 4. If Sam scored a 90 on the test, how many standard deviations from the mean is Sam’s test score?

How to read a z-table Refers to the standard normal curve (mean of 0 and standard deviation of 1) Use the table to determine the area under the curve or percentage of the area under the curve Numbers on the top and sides represent z-scores Numbers inside the table are areas/percentages

Types of areas using the z-table Areas to the left: find z-value on table EX: Given the z-score, find the area to the left under the curve. z = 0.63 EX: Given the z-score, find the probability under the normal curve for P(z < -1.45)

Types of areas using the z-table Areas to the right: find z-value on table and subtract from 1 EX: Find the area under the normal curve to the right of z = -2.72

Types of areas using the z-table Areas between two positive and two negative: find two areas and subtract (larger – smaller) EX: Find the probability under the normal curve for P (0.06 < z < 2.41)

Types of areas using the z-table Areas between one positive and one negative: subtract area to the left of the negative from the area to the left of the positive (larger – smaller) EX: Find the area under the curve between z = 0.77 and z = -0.82

Finding z-scores from data values Substitute mean, standard deviation and data value into z-score formula Simplify to get z-score EX: On a statistics test, the class mean was 63 with a standard deviation of 7. Find the area under the normal curve for a student who made a 73.

Percentiles The percent of the population that is less than or equal to a value (comparison to the rest of the data) Measures position from the minimum

Percentiles EX: A normal distribution of test scores has a mean of 83 and a standard deviation of 6. Everyone scoring at or above the 80th percentile gets placed in an advanced class. What is the cutoff score to get into the class?  EX: Suppose that you enter a fishing contest. The contest takes place in a pond where the fish lengths have a normal distribution with mean 16 inches and standard deviation 4 inches. Now suppose you want to know what length marks the bottom 10 percent of all the fish lengths in the pond. What percentile are you looking for?