Chapter 6: The Normal Distribution

Transcription

1 January 5, 2010

2 Chapter Outline 6.1 Introducing Normally Distributed Variables 6.2 Areas Under Standard Normal Distribution 6.3 Working With Any Normal Distribution 6.4 Assessing Normality: Normal Probability Plot General Objective: Introduce Normal Distribution as a continous distribution which is used extensively in Statistical Inference

3 Introduction: In Chapter 2 you learned to make histograms of continuous variables like, Height, Weight, Age, etc. The histograms are based on sample data. The histograms provide very valuable information regrading the shape of the distribution of the variable. The shape of the distribution often determines the appropriate type statistical methods for data analysis. From the shape of the histogram based on the sample data one can guess the shape of the distribution for the entire population. For example, Figure 2.8 on page 73, displays the relative-frequency histogram for the heights of 3264 female students attending a college. This histogram clearly can be approximated by a superimposed smooth curve. So one can use the smooth curve as the distribution for the entire population of college going female students. Look at the graph on the next page.

4 Both the histogram and the smooth curve show that the distribution is bell shaped. If another sample of 3264 female college students were used the histogram probably will change a little. However, the approximating smooth curve will NOT change. This smooth curve can be described by a mathematical formula which make area calculation much easier. A bell curve can be used to describe histograms in many real life problems. A Normal Distribution is described by a bell curve.

5 6.1 Normally Distributed Variable Definition 6.1 Normally Distributed variable: A variable is said to be normally distributed or to have a normal distribution if its distribution has the shape of a bell curve or a normal curve shown above.

6 A normal distribution is completely determined by its mean and standard deviation. The mean is denoted by µ and the satndard deviation is denoted by σ. µ and σ are called the parameters of the normal distribution. A normal distribution is symmetric around the mean µ. The spread is measured by σ The graph below shows three normal distributions with different µ and σ.

7 The graph of a generic normal distribution is shown below. It shows the center of the distribution is located at µ. It also shows that most of the area under the curve is between µ 3σ and µ + 3σ.

8 The sample mean and sample standard deviation of this data are µ = 64.4 and σ = 2.4. On the next page we show the superimposed normal curve with µ = 64.4 and σ = 2.4 on the relative-frequency histogram of the data in the table on left.

9 From the table on the previous page, the proportion females with between 67in and 68in = The shaded rectangle. The area under superimposed normal curve between 67in and 68in is also = The shaded area under the normal curve. This shows, in large samples, it does not matter whether the proportions is obtained from graph of a normal distribution instead of the relative-frequency table. How do we compute the area under the normal curve?

10 How do we find areas under a normal curve? Conceptually, we need a table of areas for each normal curve. However, this is impossible, because there are infinitely many normal curves - one for each µ and σ!! The way out of this difficulty is the fact that the area under any normal curve can be obtained from a standardized normal curve. If a variable X has a normal distribution with mean µ and standard deviation σ the the variable Z = X µ always has σ the same standard normal distribution no matter what the values of µ and σ are!! This fact is shown graphically on the next page. The distribution of the standardized variable Z has the mean (center) at the origin. The spread measure of Z is always 1. That is σ for the distribution of Z is 1.

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12 The graph on the left has the center at -2 and σ = 1. So only thing it needs is pushing graph to the right by 2 unit. This is done by a transformation Z = [X ( 2)]/1. The graph in the middle has the center at 3. So it needs to be shifted 3 units to the left. Also it has spread σ = 1/2. So it also needs to be stretched out by a factor of 2. This is done by a transformation Z = [X 2]/[1/2] The graph on the right has center at 9. So it needs to be shifted to the left by 9 units. It has spread σ = 3. So needs to squeezed by a factor of 3. This is done by a transformation Z = (X 9)/3 In general the transformation Z = X µ is known as σ standardizing transformation. The variable Z is called the standard normal random variable. The transformed graph is called the standard normal graph.

13 The graph on the left corresponds to mean µ and standard deviation σ. The graph on the right corresponds to mean 0 and standard deviation 1. The area between a and b is exactly same as the between z 1 = (a µ)/σ and z 2 = (b µ)/σ.

14 So, from the discussion on the previous page it is clear that we can calulate the area under any normal distribution if we can calculate the area under the standard normal ditribution. Before we try to find area under standard normal distribution, we need to know the following properties of the standard normal curve (or standard normal distribution). Property 1: The total area under the standad normal curve is 1. Property 2. The standard normal curve extends to to the left and + to the right. Property 3: The standard normal curve is symmetric around 0. This means the part of curve on the left of 0 is the mirror image of the part of the curve to the right of 0. Property 4: Almost all area is between - 4 and + 4

15 6.2 Finding Area Under Standard Normal Curve The Standard Normal Table is on page A6 and A7. (Table II). A portion of the table is reproduced here. The table always gives the area to the left of a number. The area to the right of a number or the area between two numbers have to be found by first writing it in terms of area to the left. Specific example are given on next page. z values are located on the left column and top row. The numbers in the body of the table represent the area.

16 The z-scores are located on left ( or right) column upto one decimal place. The second decimal value of z is located on the top row. For example if z= then = - ( ). To locate locate on the left column and locate n the top row. The number at the intersection of row -1.7 and column 0.05 is = The area to the left of is =

17 Example 6.3: Find the area to the left of 1.23 Draw a bell curve and locate 0 at the center on the horizontal line. Next locate the number 1.23 on the horizontal line (z axis). Shade the area that you want to find and put question mark as in the picture above. Next look at the Table II and read the number in the body of the table corresponding to row 1.2 and column This number will be

18 Example 6.4: Find the area to the right of 0.76 Draw a bell curve and locate 0 at the center on the horizontal line. Also locate the number 0.76 on the horizontal line. Shade the area that you want to find and put a question mark as in the picture above. The Table II does NOT give the area to the right of So first find the area to the left of 0.76 and write it down on the graph as shown above. The area to the right is ( 1 - area to the left) This number will be ( ) =

19 Example 6.5: Find the area between and 1.82 Draw a bell curve and locate 0 at the center on the horizontal line. Locate the number and 1.82 on the horizontal line. Shade the area that you want to find and put a question mark as in the picture above. The Table II does NOT give the area between to values. This area is the difference between the area to the left of 1.82 and the area to the left of So first find these areas and compute the difference. The area between and 1.82 = =0.7173

20 Example 6.6: The z-scoe having a specified area to its left Draw a bell curve and shade the given area. Pay attention to left or right tail. Point an arrow to the z-value that you want to find and put a question mark as in the picture above. Locate the given area in the body of the Table II. Then read the corresponding z-value from the z-column on the left and top row. For this example z-value corresponding 0.04 is

21 Definition 6.3 The z α notation. The symbol z α is used to denote the z-score that has an area α to its right under the standard normal curve. See the graph below.

22 Finding z α : Use Table II to find: (a) z 0.025, (b) z 0.05 (a) z is the z-score that has an area of to its right. Draw a bell curve, shade ar area of at the right tail of the bell curve. Indicate the z-score by z α and put a question mark to remind yourself that this is the value you are looking for. The area to the right is means the area to its left is Now in the body of Table II locate the number closest to and read the corresponding z-value. This value is write this number next to z as in the graph above.

23 (b) Follow the same steps for z The z 0.05 = The area to the right is 0.05 means the area to its left is 0.95 In the body of Table II, the numbers that are close to 0.95 are and The corresponding z values are 1.64 and Since 0.95 is between and , take the average of 1.64 and 1.65 as z-value corresponding to 0.95

24 In the graph above the central area is 0.95 Draw the graph of a bell curve. Locate the central area and mark -z and z value on the horizontal axis. Put a question mark to indicate that this is what you are looking for. The central area 0.95 means the two tails together is Since the bell curve is symmetric, this means each tail area must be 0.05/2 =025. Locate this on the graph. This gives the area to the left of z = = Now use Table II find the z-value with area to its left = This value is 1.96

25 6.3 Working With Any Normal Distribution: So far we computed various types of areas under standard normal graph. We also computed z values for given area of different types. We need to compute various types of areas under any normal (rather than only standard normal) We also need to compute X value for given areas under normal with any mean and any standard deviation.

26 Procedure 6.1 Determine areas under normal distribution with mean µ and standard deviation σ. Step 1: Draw a bell curve. Mark the center with µ on the horizontal axis. Step 2:Mark the x-values on the horizontal axis and shade the region whose area you want to compute. Step 3:Transform the delimiting x values (denoted by a nd b in the graph) to corresponding z-scores. z-score for a is z 1 = z = a µ. Similarly z-score for b is σ z 2 = z = b µ. mark these values right under a and b. σ Now using Table II compute the area between z 1 and z 2.

27 Example 6.9 Computing Percentages for N(µ, σ) IQ scores are known to have a normal distribution with mean µ = 100 and standard deviation σ = 16. Determine the percentage of people who have IQs between 115 and 140. Solution: See the graph on the next page. Step 1: Draw a bell curve and mark the center with µ = 100. Step 2: Mark the values a=115 aand b=140 on the horizontal X-axis. Shade the region between 115 and 140 Step 3: Compute z-score for 115: z = ( )/16 = 0.94 Step 4: Compute z-score for 140: z = ( )/16 = 2.50 Step 5: Use Table II to compute the area between 0.94 and This area is = = Step 6: Interpretation: 16.74% people have IQ between 115 and 140.

28 Figure 6.19

29 The Rule Property 1: 68.26% of all possible observations lie within µ σ and µ + σ. Property 2: 95.44% of all possible observations lie within µ 2σ and µ + 2σ. Property 3: 99.74% of all possible observations lie within µ 3σ and µ + 3σ.

30 Example 6.10 The Rule: Apply this rule to IQ. Compute the (µ σ, µ + σ), (µ 2σ, µ + 2σ) and (µ 3σ, µ + 3σ) ranges for IQ score. Solution: For IQ scores µ = 100 and σ = 16. Hence One sigma range: µ σ = = 84 and µ + σ = = 116 two sigma range: µ 2σ = = 68 and µ + 2σ = = 132 three sigma range: µ 3σ = = 52 and µ + 3σ = = 148

31 Procedure 6.2 Determine X-score (raw score) for a given probability. Step 1: Draw a bell curve. Step 2: Shade the region of interest and write down the associated probability. Step 3: Use Table II to calculate the corresponding z-score Step 4: Use the formula X = µ + zσ to calculate X-score or raw score.

32 Example 6.11 Determine X-score (raw score) for a given probability. Determine the 90 th percentile of the IQ scores. Solution: 90th Percentile of IQ is a value X 90 such that 90% of the people have the score below X 90. This means the area to the left of X 90 should be The 90 th percentile for Z distribution is z 90 = 1.28 (using Table II). Hence X 90 = µ + zσ = (16) = = 120.8

33 Computing Area Under Normal Curve Using StatCrunch Computing Area under any normal curve for given X value is very easy in StatCrunch. Computing X value under any normal curve for given Area is also very easy in StatCrunch. NOTE: StatCrunch can calculate Left Tail area as well as Right Tail area. Remember Table II only gives the left tail area. However, before using StatCrunch, draw a bell curve and indicate what exactly you want to compute. Specifically, decide whether you want to compute area for a specified z or X score or you want to compute z or X score for a specified area. This is why you need to draw the bell curve by hand and indicate what you are looking for. The next slide explains the components of StatCrunch screen.

34 Compnents of StatCrunch Screen for computing Area Follow: Stat > Calculator > Normal. Screen below will appear. Enter the value of Mean and Standard deviation. In the first box on the second row, you have two choices: <= for left tail or => for right tail. Select the appropriate one. Next two boxes are for X and Area. However, you need to enter only one. To compute area, enter X (middle box) and press Compute To compute X, enter area (last box) and press Compute

35 Simulating Normal Curve Using StatCrunch OutPut 6.1 Simulating Normally Distributed Variable Human Gestation Periods are normally distributed with mean µ = 266 and standard deviation σ = 16 days. Simulate 1000 gestation periods. Make a histogram of the simulated data with a superimposed normal curve. StatCrunch instruction on the next page.

36 StatCrunch Steps: Open StatCrunch and select Chapter 6 Follow Data> Simulate Data > Normal You should see the following screen.

37 Enter 1000 for Rows box. Enter 1 for column, Enter 266 for mean and enter 16 for standard deviation. Now press the tab Simulate. This will show the simulated data in a column with heading Normal1.

38 Now Follow: Graphics > Histogram. You should the screen below.

39 Now Select Column: Normal1. The column name should appear on the box on the right. You should the screen below. Now press Next.

40 In box neext to Type select, Density. In the box next to Binwidth select a number. I selected 2. You can try with other Binwidth. Press Next

41 In the Overlay Density box, select Normal. Enter the mean and standard deviation. Now press Create Graph.

42 Now copy this Histogram to your assignment file.