Chapter 6. THE NORMAL DISTRIBUTION Introducing Normally Distributed Variables The distributions of some variables like thickness of the eggshell, serum cholesterol concentration in blood, white blood cells count in a specimen of blood have roughly the shape of a normal curve (bell shaped curve) 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 normal curve. Normal distribution (curve) completely determined by mean ( ) and standard deviation ( ). Parameters of Normal distribution = (, ) Characteristics of Normal distribution Bell-shaped Symmetric around the mean Close to the horizontal axis outside the range from -3 to +3 Spread depends on the standard deviation. Area under the curve is 1 for any (, ). Notation: Y~N(12, 7) indicates that Y has normal distribution with mean 12 and standard deviation 7 1
Normally distributed variables and normal-curve areas For a normally distributed variable, the percentage of all possible observations that lie within any specified range equals the corresponding area under its associated normal curve expressed as a percentage. This holds true approximately for a variable that is approximately normally distributed. Example (Heights of Female of College Students): A college has an enrollment of 3264 female students. Records show that the mean height of these students is 64.4 inches and the standard deviation is 2.4 inches. Since the shape of the relative histogram of this sample college students approximately normally distributed, we assume the total population distribution of the height of all the female college students follows the normal distribution with the same mean and the standard deviation. Now if you want to find out the percentage of students whose heights are between 66 and 68 inches, you have to evaluate the area under the normal curve from 66 to 68. 2
68 Area = 66 1 2 (2.4) 2 e 2 ( x 64.4) 2 2( 2.4) dx = 0.1846 (by TABLE) Relative frequency = 0.1100+0.0735 = 0.1835 (by relative frequency distribution) Standardizing a Normally Distributed Variable Facts: 1) Once we know the mean and standard deviation of a normally distributed variable, we know its distribution and associated normal curve 2) Percentages for a normally distributed variable are equal to areas under its associated normal curve. How do we find areas under a normal curve? Integration? Or tables for each different and? Or standardize your normal curve and use only one table with mean( )=0 and standard deviation( )=1? Standard Normal Distribution; Standard Normal Curve A normally distributed variable having mean 0 and standard deviation 1 is said to have the standard normal distribution. Its associated normal curve is called the standard normal curve. Standardized Normally Distributed Variable The standardized version of a normally distributed variable Y, Z= Y μ σ standard normal distribution. has the 3
Areas under the Standard Normal Curve Basic Properties of the Standard Normal Curve 1. The total area under the standard normal curve is equal to 1. 2. The standard normal curve extends indefinitely in both directions, approaching, but never touching, the horizontal axis as it does so. 3. The standard normal curve is symmetric about 0; i.e., the part of the curve to the left of 0 is the mirror image of the part of the curve to the right of 0. 4. Most of the area under the standard normal curve lies between 3 and 3. Using the Standard-Normal Table There are infinitely many normally distributed variables, however, if these variables can be standardized, then the standard normal tables can be used to find the areas under the curve. * Table set up to accumulate the area under the curve from - to and specified value. * The table starts at 3.9 and goes to 3.9 since outside this range of values the area is negligible. * The table can be used to find a z value given and area, or and area given a z value. 4
The z Notation The symbol z is used to denote the z- score having area (alpha) to its right under the standard normal curve. z - z sub alpha or simply z. Working with Normally Distributed Variables To Determine a Percentage or Probability for a normally Distributed Variable 1. Sketch the normal curve associated with the variable. 2. Shade the region of interest and mark the delimiting x-values. 3. Compute the z-scores for the delimiting x-values found in step 2. 4. Use Table II to obtain the area under the standard normal curve delimited by the z-scores found in step 3. Example (contd.) Height of Female students: Normal distribution with = 64.4, = 2.4. We want to determine the probability that randomly selected student will have height between 66 and 68. z-score for x = 66: z = (66-64.4)/2.4 = 0.67, x=68: z = (68-64.4)/2.4 = 1.5 area under standard normal curve: z= 1.5 -> 0.9332, z = 0.67 -> 0.7486 resulting probability: 0.9332 0.7486 = 0.1846 In conclusion: For normally distributed variables Y questions: 1)What percentage of values of Y are in the range a to b 2)For randomly selected Y what is the probability P(a< Y < b) can both be answered by computing area under the normal curve between a and b. 5
Visualizing a Normal Distribution 1. 68.26% of all possible observations lie within one standard deviation to either side of the mean, i.e., between - and +. 2. 95.44% of all possible observations lie within two standard deviations to either side of the mean, i.e., between - 2 and + 2. 3. 99.74% of all possible observations lie within three standard deviations to either side of the mean, i.e., between - 3 and + 3. To Determine the Observations Corresponding to a specified Percentage or Probability for a Normally Distributed Variable. 1. Sketch the normal curve associated with the variable. 2. Shade the region of interest. 3. Use Table II to obtain the z-scores delimiting the region in step 2. 4. Obtain the x-values having the z-scores found in step 3: x=μ+ z(σ) Example (contd.) a. Obtain the Q 3 (75 th percentile) of the height of female students. The z-score corresponding to Q 3 is the one having an area of 0.75 to its left under the standard normal curve. From Table II, that z-score is 0.67, approximately. So the x-value (height) corresponding to that z-score is 64.4 + (0.67)*2.4 = 66 inches. b. Obtain the 10 th percentile. z-score corresponding to P 10 is the one having an area of 0.1 to its left under the standard normal curve. From Table II, that z-score is 1.28, approximately. So the x-value (height) corresponding to that z-score is 64.4 + (-1.28)*2.4 = 61.32. Assessing Normality. Normal Probability Plots. 1. Many statistical procedures are based on the assumption that data analyzed is coming from normally distributed populations. One way to assess the normality of your data is through the use of Empirical Rule. We can compute percentages within 1, 2 and 3 SD-s from the mean of the data and check is the percentages are close to expected 68-95-99.7. Visual check of the histogram is also helpful, if we have unimodal, nearly symmetric graph with no long or very short tails, we can be pretty sure that normality assumption can be made. 2. With small data sets in particular Empirical Rule or a visual check of the histogram is not as useful. A special statistical graph: Normal Probability Plot is often used. The plot is a scatterplot that compares observed data values to the values we would expect to have if the population were normal. If the data came from normal population, points would follow a straight line; Following example illustrates the procedure. 6
Ex. Y= age of onset of diabetes, sample of size 5: 7, 48, 43, 51, 49. Order your data. Compute mean and standard deviation: y=39.6, s=18.46 Y=observed height 7 43 48 49 51 (i-.5)/n= adjusted percentile.10.30.50.7.9 z α -1.28-0.52 0 0.52 1.28 Y=theoretical height= 16.0 30.0 39.6 49.2 63.2 y+ z α s Graphing theoretical height (x-axis) vs observed height (y-axis) we can see that points do not follow a line, first value is much smaller than expected theoretical value indicating left skewness. Following two pictures illustrate Normal Quantile plots for other data sets. normal Ex1 Rainwater ph value Good fit, distribution is close to population normal have particularly Ex2 Survival times in some Curved pattern, distribution is not but right skewed, few individuals long survival times. 7
Interpreting Normal plots: Distribution close to normal will be indicated by a straight line (more or less) Left skewed distribution will be indicated by the line curving down from the left (lower observations smaller than expected for normal distr.). Right skewed distribution will be indicated by the line curving up from the right (upper observations larger than expected for normal distr.) Outliers will appear as points far away from the overall pattern of the plot. Granularity: sometimes plotted points appear to form a horizontal segments indicating repeated identical observations. This should not prevent us from adopting a normal distribution for the data. We can avoid this problem often by taking more precise measurements (not rounding to much). Instructions for : TI83, 83-Plus, 84-Plus Computing areas under normal curves: use 2 nd VARS to get to the DISTR menu: option 2 normalcdf(lower limit, upper limit, mean, standard deviation) will give are between lower and upper limits (mean=0 and SD=1 are default values) Ex1 To find area between 1 and 1.7 under N(0,1) use normalcdf(1,1.7,0,1)=.1141 Ex2 To find area onder N(0,1) left of 2.3 use normalcdf(-1000000, 2.3,0,1)=.9893 (use any large negative number as lower limit) Ex3 To find area right of 2.11 under N(0,1) use normalcdf(2.11,1000000,0,1)=.0174 (use any large positive number as upper limit) Ex4 To find area under curve N(12,3) between 10 and 16 use normalcdf(10, 16, 12,3)=.6563 Finding points from under the normal curves when area is given. use 2 nd VARS to get to the DISTR menu: option 3 invnorm(area to the left, mean, standard deviation) (mean=0 and SD=1 are default values) 8
Ex5 To find third decile of N(0,1) use invnorm(.3,0,1)=-.52 Ex6 To find 95 th percentile of N(0,1) (or to find Z. 05 ) use invnorm(.95,0,1)=1.645 Ex7 To find third quartile on N(12,3) use invnorm(.75,12,3)=14.02 9