Mat231 Probability And Statistics Ii Lecture 1 PDF

MAT231 Probability and Statistics II Lecture 1 Normal Distribution Lecturer: Prof. Emad Ashmawy 1  Assessment Criteria 25 Marks: Mid-term Exam 10 Marks: Tutorial Quizzes 10 Marks: Assignments 15 Marks: Lab and Lab Exam 40 Ma...

MAT231 Probability and Statistics II Lecture 1 Normal Distribution Lecturer: Prof. Emad Ashmawy 1  Assessment Criteria 25 Marks: Mid-term Exam 10 Marks: Tutorial Quizzes 10 Marks: Assignments 15 Marks: Lab and Lab Exam 40 Marks: Final exam 2  Contents / weeks week topic 1 Normal Probability Distribution 2 Normal Probability Distribution (continue) 3 Sampling Methods and Central Limit Theorem 4 Sampling Methods and Central Limit Theorem (continue) 5 Distributions derived from normal distribution (Chi-squared, student-t. and F) 6 Statistical estimation, point estimation and confidence intervals 7 Mid-term Exam 8 One sample tests of hypotheses 9 One sample tests of hypotheses 10 One sample tests of hypotheses 11 Two sample tests of hypotheses 12 Two sample tests of hypotheses 13 Linear Regression and Correlation 14 Analysis of variance 15 Revision 3 Characteristics of a Normal Probability Distribution 1. It is bell-shaped and has a single peak. 2. It is symmetrical about the mean. 3. It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. 4. The arithmetic mean, median, and mode are equal 5. The total area under the curve is 1.00. 6. The area to the left of the mean = area right of mean = 0.5. 4 The Normal Distribution – Graphically 5 The Family of Normal Distribution Equal Means and Different Standard Deviations 6 The Family of Normal Distribution Different Means and Standard Deviations 7 The Family of Normal Distribution Different Means and Equal Standard Deviations 8 The Standard Normal Probability Distribution  The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.  It is also called the z distribution.  A z-value is the signed distance between a selected value, designated X, and the population mean , divided by the population standard deviation, σ.  The formula is: 9 10  Areas Under the Normal Curve 11  The Normal Distribution – Example The weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the z-value for the income, let’s call it X, of a foreman who earns $1,100 per week? For a foreman who earns $900 per week? 12  Normal Distribution – Finding Probabilities In an earlier example, we reported that the mean weekly income of a shift foreman in the glass industry is normally distributed with a mean of $1,000 and a standard deviation of $100. What is the likelihood of selecting a foreman whose weekly income is between $1,000 and $1,100? 13  Normal Distribution – Finding Probabilities 14  Normal Distribution – Finding Probabilities Using the Normal Distribution Table 15  Finding Areas for z Using Excel The Excel function =NORMDIST(x,Mean,Standard_dev,Cumu) =NORMDIST(1100,1000,100,true) generates area (probability) from Z=1 and below 16 Normal Distribution – Finding Probabilities (Example 2) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $790 and $1,000? Excel Function: =NORMDIST(1000,1000,100,true)-NORMDIST(790,1000,100,true) 17  Normal Distribution – Finding Probabilities using the Normal Distribution Table 18 Normal Distribution – Finding Probabilities (Example 3) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Less than $790? 19 Excel Function: =NORMDIST(790,1000,100,true)  Normal Distribution – Finding Probabilities Using the Normal Distribution Table 20  Normal Distribution – Finding Probabilities (Example 4) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $840 and $1,200? 21 Excel Function: =NORMSDIST(2.0)-NORMSDIST(-1.6)  Normal Distribution – Finding Probabilities Using the Normal Distribution Table 22  Normal Distribution – Finding Probabilities (Example 5) Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is: Between $1,150 and $1,250 23 Excel Function: =NORMSDIST(2.5)-NORMSDIST(1.5)  Normal Distribution – Finding Probabilities Using the Normal Distribution Table 24 25 26