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1. Standard Normal Distribution Questions There were three types of standard normal distribution questions in the 5th quiz. 1) Probability that Z is smaller than a specified value 2) Probability that Z is larger than a specified value 3) Probability that Z is between two specified values . . . Standard normal distribution has a mean value of zero and a standard deviation value of 1. Its shape is given in the following graph. Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.2 0.1 0.0 -3 -2 -1 0 X 1 2 3 All three statements form an area under this curve. Usually, in order to find the area under the curve, we use integrals. However, the integral of the normal distribution function does not have a simple algebraic form. Therefore, z values are tabulated and the corresponding probabilities are given in standard normal distribution tables. These tables generally have the first digit in the rows, and the second digit in the columns, and inside the table we see the probability values. An example of standard normal distribution table is given below. In the rows, we see numbers .0, .1, .2, ... and in the columns we see numbers .00, .01, .02, ... For example, in order to find the value that corresponds to 1.13 we first find 1.1 in the rows, and then we find 0.03 in the columns (note that 1.1+0.03=1.13). This value is the probability that z is smaller than 1.13. Formally, Instead, if we want to calculate the probability that z is larger than 1.13, we make use of the fact that the total area under the curve equals 1. Therefore, When z values are negative, we make use of the fact that the normal distribution table is symmetrical. For example, the probability that z is larger than 1.13 is equal to the probability that z is smaller than -1.7. See the graphs for better understanding. Distribution Plot Normal, Mean=0, StDev=1 0.4 Density 0.3 0.2 P(z>1.13) P(z<-1.13) 0.1 0.1292 0.0 0.1292 -1.13 0 X It is also valid when the inequality signs change. For example, 1.13 Distribution Plot Normal, Mean=0, StDev=1 0.4 0.3 Density 0.871 0.2 P(z<1.13) 0.1 0.0 0 X 1.13 Distribution Plot Normal, Mean=0, StDev=1 0.4 0.3 Density 0.871 0.2 P(z>-1.13) 0.1 0.0 -1.13 0 X For the area between two specified values, we use subtraction. For example, if we want to find the area between -1.13 and 1.13 we use To sum up, 1) Probability that Z is smaller than a specified value . o If z is positive, find the value that corresponds to z in the table. Example: o If z is negative, find the value that corresponds to |z| in the table, and subtract it from 1. Example: 2) Probability that Z is larger than a specified value . o If z is positive, find the value that corresponds to z in the table, and subtract it from 1. Example: o If z is negative, find the value that corresponds to |z| in the table. Example: 3) Probability that Z is between two specified values . o Find Example: 2. Normal Distribution Questions For normal distribution questions, we first find the corresponding z values, and then do the calculations that we do in standard normal distribution questions. For example, - Given that X follows a normal distribution with mean 52 and standard deviation 5, what is P(50<X<57)? We first find the corresponding z values for 50 and 57. Now that we have transformed our normal random variables to standard normal variables, we can find the corresponding probabilities. First we find from the table (Row: 1.0, Column: .00) as 0.8413. Then, we find using the value corresponds to 0.4 (Row: .4, Column: .00) as 0.6554 and subtract it from 1 (1-0.6554=0.3446). Therefore, 3. Uniform Distribution Questions There were 2 uniform distribution questions. The first one asked the probability density function value, and the second one asked the expected value. X follows a continuous uniform distribution with a minimum value of 63 and a maximum value of 74. What is the probability density function value at X=69.96? First, let’s remember the shape of continuous uniform distribution. Distribution Plot Uniform, Lower=63, Upper=74 0.09 0.08 0.07 Density 0.06 0.05 0.04 0.03 0.02 0.01 0.00 62 64 66 68 X 70 72 74 It has a rectangular shape which makes the calculations very easy. The minimum value is 63 and the maximum value is 74. We also know that the area of this rectangle is equal to 1. Therefore, in order to find the height of the rectangle (which is equal to the probability density function value) we use the following formula: Note that the probability density function value for all possible x values are the same. Therefore, it is not important if the problem asks the pdf value at X=69.96 or X=73.81 or X=65.42. They are all the same and equal to 0.09. The second uniform distribution question was about the expected value. X follows a uniform distribution with a minimum value of 31.6 and a maximum value of 54.4. What is the expected value of X? Again, since the shape of this distribution is rectangular, the calculations are very easy. The expected value is the average value and all the points have the same likelihood of occurrence. Therefore, the expected value is the middle point (between maximum and minimum values). 4. Matching Questions 5. Poisson and Binomial Distribution Questions There were two questions for these distributions. When an unfair coin tossed, it land tails 80% of the time. What is the probability that when it's tossed 23 times, it will land heads at most twice? This is clearly a binomial distribution question with parameters p=0.2, n=23 and k=2. You are expected to find . Average number of blackouts is 48 per year in a city. What is the probability that there will be less than or equal to 2 blackouts in a month? Again, this is clearly a Poisson distribution question with parameter =48 blackouts/year. You are expected to find in this question as well. Refer to the solution key of the 4th quiz for these questions. Since I gave detailed explanations there, I will not solve them again.