6.2 Areas Under the Standard Normal N(0,1) Curve. 1. The total area under the N(0,1) 3. Most of the area under N(0,1) curve lies between

Transcription

1 6.2 Areas Under the Standard Normal N(0,1) Curve Basic Properties: 1. The total area under the N(0,1) 2. The N(0,1) is symmetric 3. Most of the area under N(0,1) curve lies between

2 Using the standard normal table: Ex: Determine the area under the N(0,1) curve that a. lies to the left of P(Z 2.11) 124

3 b. lies to the right of P (Z -1.25) 125

4 c. lies between -0.5 and 2.47 inclusive P(-0.5 Z 2.47) 126

5 Finding the z-score for a specified area: Ex. Determine the z-score having an area of a to its left There is no 0.05 in table, but Hence (using interpolation) The z-score is 127

6 b to its right Same as: Area to its left is From table: Hence, z-score is 128

7 Notation: z α denotes the z-score having area (alpha) to its right under N(0,1) curve. From above : z What is z 0.05? z

8 6.3 Working with Normally Distributed Variables To determine a percentage or probability for a normally dist ed variable: Steps: 1. Sketch the normal curve. 2. Shade the region of interest and mark delimiting x-values. 3. Compute the z-scores for the delimiting x-values found in (2). 4. Use table II to obtain the area under the N(0,1) curve. 130

9 Ex. Each year, thousands of college seniors take the Graduate Record Examination (GRE). The scores are transformed so that they have a mean of 500 and a SD of 100. Furthermore, the scores are known to be normally dist ed. Determine the percentage of students that score: a. between 350 and 600 inclusive. Step 1 & 2: pdf Score 131

10 Step 3: P(350 X 600)? (Shaded area). To use Normal tables, first transform the normal RV X into the standard normal RV Z (find z-scores): or Step 4: P(350 X 600) (-1.5 Z 1) Represents the area under N(0,1) over the interval from -1.5 to

11 b. 375 or grater pdf Score P(X 375) P ( Z ) 133

12 c. between 300 and pdf Score P(300 X 450) d. exactly equal to 680. P(X 680) 134

13 e. What score is exceeded by exactly 5%? (95 percentile) pdf Score P(Z < a) 0.95 a Excel: x 135

14 6.4 Assessing Normality; Normal Probability Plots If variable is normally dist ed, to assess normality: Large sample look at histogram. Small/large look at normal sample probability plots. Normal probability plot: If plot is roughly linear, then accept as reasonable that the variable is Note : You do not need to know the mechanics of how to construct a normal probability plot. 136

15 THE SAMPLING DISTRIBUTION OF THE MEAN Chapter Sampling Error; The Need for Sampling Dist ns Sampling error is the error resulting from using a Recall: Descriptive measures of a population are called parameters. Example: Descriptive measures calculated from a sample are called Example: 137

16 The distribution of a statistic is called the Questions: - What is the dist'n of - What is the dist'n of 7.3 The Sampling Distribution of Mean Idea: From repeated sampling Sample 1 of size n Sample 2 of size n Sample m of size n Questions: What is the? What is the? What is the? 138

17 Central Limit Theorem (CLT) If relatively large samples of size n are drawn from any population, the sampling dist'n of X is approximately normal. - If the popul'n dist'n is normal, the sampling dist'n of X will be - If the population dist'n is non-normal, the sampling dist'n of X will be Mean and SD of X? 139

18 7.2 The Mean and SD of X Formula 7.1 The mean of X, for samples of size n, is equal to the mean of the original popul'n. That is: µ X Formula 7.2 The SD for X, for samples of size n, equals the SD of the parent popul n divided by the square root of the sample size. σ X The SD of a statistics is called the standard error of the statisic. σ X : is called the 140

19 Example 1 The amount of sulfur in the daily emissions from a power plant has a normal dist n with a mean of 94 pounds and a SD of 22 pounds. a. What is the mean and SD of the parent popul'n? X : µ σ 141

20 b. If 5 days are randomly selected and the average sulfur emission is calculated, what is the mean and SD of the sample mean? Mean: µ X SD: σ X X ~ N( ) 142

21 c. Plot the two distributions found in (a) original population and (b) average of 5 observations Sulfur Emissions - Pounds 143

22 d. Find the probability that on a randomly chosen day, sulfur emissions are more than 100 pounds. P[X > 100]? Sulfur Emissions - Pounds P [ X > 100] P [ Z > ] P [ Z > ] 144

23 e. Find the probability that, if five days are randomly selected, their mean emission exceeds 100 pounds. P[ X > 100]? Sulfur Emissions - Pounds In this case, the transformation needed to standardized the normal RV is: Z X - σ µ x X 145

24 P [X > 100 ] P [ Z > ] 146

25 Example 2 The mean and SD of the strength of a packaging material are 55 and 7 pounds, respectively. If 45 specimens of this material are tested, a. Is it reasonable to assume a normal dist n for the sample mean X? Why or why not? X ~ N(, ) 147

26 b. Find the probability that the sample mean strength X will be between 54 and 56 pounds? P[54 < X < 56]? 148