Introduction Previous lessons demonstrated the use of the standard normal distribution. While distributions with a mean of 0 and a standard deviation of 1 are rare in the real world, there is a formula that allows us to use the properties of a standard normal distribution for any normally distributed data. With this formula, we can generate a number called a z-score to use with our data. This makes the normal distribution a powerful tool for analyzing a wide variety of situations in business and industry as well as the physical and social sciences : Standard Normal Calculations

Introduction, continued Using and understanding z-scores requires a deeper understanding of standard deviation. In the previous sub-lesson, we found the standard deviations of small data sets. In this lesson, we will explore how to use z-scores and graphing calculators to evaluate large data sets : Standard Normal Calculations

Key Concepts Recall that a population is all of the people or things of interest in a given study, and that a sample is a subset (or smaller portion) of the population. Samples are used when it is impractical or inefficient to measure an entire population. Sample statistics are often used to estimate measures of the population (parameters). The mean of a sample is the sum of the data points in the sample divided by the number of data points, and is denoted by the Greek letter mu, μ : Standard Normal Calculations

Key Concepts, continued The mean is given by the formula, where each x-value is a data point and n is the total number of data points in the set. From a visual perspective, the mean is the balancing point of a distribution. The mean of a symmetric distribution is also the median of the distribution. The median is the middle value in a list of numbers. Both the mean and median are at the center of a symmetric distribution : Standard Normal Calculations

Key Concepts, continued The standard deviation of a distribution is a measure of variation. Another way to think of standard deviation is “average distance from the mean.” The formula for the standard deviation is given by, where (the lowercase Greek letter sigma) represents the standard deviation, x i is a data point, and means to take the sum from 1 to n data points : Standard Normal Calculations

Key Concepts, continued Summation notation is used in the formula for calculating standard deviation; it is a symbolic way to represent the sum of a sequence. Summation notation uses the uppercase version of the Greek letter sigma, Σ. After calculating the standard deviation, σ, you can use this value to calculate a z-score. A z-score measures the number of standard deviations that a given score lies above or below the mean. For example, if a value is three standard deviations above the mean, its z-score is : Standard Normal Calculations

Key Concepts, continued A positive z-score corresponds to an individual score that lies above the mean, while a negative z-score corresponds to an individual score that lies below the mean. By using z-scores, probabilities associated with the standard normal distribution (mean = 0, standard deviation = 1) can be used for any non-standard normal distribution (mean ≠ 0, standard deviation ≠ 1). The formula for calculating the z-score is given by, where z is the z-score, x is the data point, μ is the mean, and σ is the standard deviation : Standard Normal Calculations

Key Concepts, continued z-scores can be looked up in a table to determine the associated area or probability. The numerical value of a z-score can be rounded to the nearest hundredth. Graphing calculators can greatly simplify the process of finding statistics and probabilities associated with normal distributions : Standard Normal Calculations

Common Errors/Misconceptions calculating and applying a z-score to a distribution that is not normally distributed using the area to the left of the z-score when the area to the right of the z-score is the area of interest and vice versa misreading the table with the associated probability : Standard Normal Calculations

Guided Practice Example 1 In the 2012 Olympics, the mean finishing time for the men’s 100-meter dash finals was seconds and the standard deviation was 0.72 second. Usain Bolt won the gold medal, with a time of 9.63 seconds. Assume a normal distribution. What was Usain Bolt’s z-score? : Standard Normal Calculations

: Standard Normal Calculations Guided Practice: Example 1, continued 1.Write the known information about the distribution. Let x represent Usain Bolt’s time in seconds. μ = = 0.72 x = 9.63

Guided Practice: Example 1, continued 2.Substitute these values into the formula for calculating z-scores. The z-score formula is Usain Bolt’s z-score for the race was –0.65. Therefore, his time was 0.65 standard deviations below the mean : Standard Normal Calculations ✔

Guided Practice: Example 1, continued : Standard Normal Calculations

Guided Practice Example 2 What percent of the values in a normal distribution are more than 1.2 standard deviations above the mean? : Standard Normal Calculations

: Standard Normal Calculations Guided Practice: Example 2, continued 1.Sketch a normal curve and shade the area that corresponds to the given information. Start by drawing a number line. Be sure to include the range of values –3 to 3. Create a vertical line at 1.2. Shade the region to the right of 1.2.

Guided Practice: Example 2, continued 2.Use a table of z-scores or a graphing calculator to determine the shaded area. A z-score table can be used to determine the area. Since the area of interest is 1.2 standard deviations above the mean and greater, we need to look up the area associated with a z-score of 1.2. The following table contains z-scores for values around 1.2σ : Standard Normal Calculations

Guided Practice: Example 2, continued : Standard Normal Calculations z

Guided Practice: Example 2, continued To find the area to the left of 1.2, locate 1.2 in the left-hand column of the z-score table, then locate the remaining digit 0 as 0.00 in the top row. The entry opposite 1.2 and under 0.00 is ; therefore, the area to the left of a z-score of 1.2 is or 88.49%. We are interested in the area to the right of the z- score. Therefore, subtract the area found in the table from the total area under the normal distribution, 1. 1 – = The area greater than 1.2 standard deviations under the normal curve is about or 11.51% : Standard Normal Calculations

Guided Practice: Example 2, continued Alternately, you can use a graphing calculator to determine the area of the shaded region. Note: The lower bound is 1.2, but the upper bound is infinity, so any large positive integer will work as the upper bound value. Use 100 as the upper bound. Since this problem is based on standard deviations under the standard normal distribution, the mean = 0 and the standard deviation = : Standard Normal Calculations

Guided Practice: Example 2, continued On a TI-83/84: Step 1: Press [2ND][VARS] to bring up the distribution menu. Step 2: Arrow down to 2: normalcdf. Press [ENTER]. Step 3: Enter the following values for the lower bound, upper bound, mean (m), and standard deviation (σ). Press [ENTER] after typing each value to navigate between fields. Lower: [1.2]; upper: [100]; μ: [0]; σ: [1]. Step 4: Press [ENTER] twice to calculate the area of the shaded region : Standard Normal Calculations

Guided Practice: Example 2, continued On a TI-Nspire: Step 1: Press the [home] key. Step 2: Arrow over to the spreadsheet icon and press [enter]. Step 3: Press the [menu] key. Arrow down to 4: Statistics, then arrow right to bring up the sub-menu. Arrow down to 2: Distributions and press [enter]. Step 4: Arrow down to 2: Normal Cdf. Press [enter] : Standard Normal Calculations

Guided Practice: Example 2, continued Step 5: Enter the values for the lower bound, upper bound, mean (μ), and standard deviation (σ), using the [tab] key to navigate between fields. Lower Bound: [1.2]; Upper Bound: [100]; μ; [0]; σ: [1]. Tab down to “OK” and press [enter]. Step 6: The values entered will appear in the spreadsheet. Press [enter] again to calculate the area of the shaded region. The area returned on either calculator is about or 11.51% : Standard Normal Calculations ✔

Guided Practice: Example 2, continued : Standard Normal Calculations

Example 3 If a population of human body temperatures is normally distributed with a mean of 98.2ºF and a standard deviation of 0.7ºF, estimate the percent of temperatures between 98.0ºF and 99.0ºF : Standard Normal Calculations

1 - Write the known information: μ = 98.2 = 0.7 x(lower) = 98 x(upper) = : Standard Normal Calculations

2 – Sketch it : Standard Normal Calculations The shaded area is the percent of data we are trying to estimate.

: Standard Normal Calculations Z-score Table lookup: What is the percent of data to the left of this upperbound z-score?

: Standard Normal Calculations Z-score Table lookup: What is the percent of data to the left of this lowerbound z-score?

With the percentage to the left of the upperbound z- score and the percentage found from the lowerbound z- score, how do we estimate the percentage between 98.0 and 99.0? : Standard Normal Calculations UpperboundLowerbound