Continuous Random Variable So how do we determine 𝑓(𝑥), the probability density function of 𝑋? Choose pdf from a class of continuous distributions Selection depends on the nature of population of interest Certain types of data have known distributions Otherwise, make an assumption or educated guess in choosing the most appropriate one In this class, it should always be clear what the distribution of something is.

Normal Distribution The mean, median, and mode are all located Most prevalent and arguably the most important class of distributions Bell curve Symmetric about 𝜇 PDF: 𝑓 𝑥 = 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 ) 𝑥= value of 𝑋 𝜇=𝐸(𝑋) (population mean of 𝑋) 𝜎 2 =𝑉𝑎𝑟(𝑋) (population variance of 𝑋)  𝜎= 𝑉𝑎𝑟(𝑋) (population standard deviation of 𝑋) 𝜋=3.14159… (pi) 𝑒=2.71828… (natural number) The mean, median, and mode are all located at the same place.

Normal Distribution Most prevalent and arguably the most important class of distributions Bell curve Symmetric about 𝜇 PDF: 𝑓 𝑥 = 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 ) 𝑥= value of 𝑋 𝜇=𝐸(𝑋) (population mean of 𝑋) 𝜎 2 =𝑉𝑎𝑟(𝑋) (population variance of 𝑋)  𝜎= 𝑉𝑎𝑟(𝑋) (population standard deviation of 𝑋) 𝜋=3.14159… (pi) 𝑒=2.71828… (natural number) 𝑃 𝑋<𝑎 = −∞ 𝑎 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 ) 𝑑𝑥 = ? a Draw this on the board. Interpretation of the standard deviation for normal distribution: 68-95-99.7 rule 𝑃(𝜇−𝜎≤𝑥≤𝜇+𝜎)≈0.6827 𝑃(𝜇−2𝜎≤𝑥≤𝜇+2𝜎)≈0.9545 𝑃(𝜇−3𝜎≤𝑥≤𝜇+3𝜎)≈0.9973 That is, about 68% of observations lie within one SD of the mean, about 95% of observations lie within two SDs of the mean, and about 99.7% of observations lie within three SDs of the mean.

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? c

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! c

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 0.1587 1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 0.8413 1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 𝑃 𝑍≤−1 =𝑃 𝑍≥1 Because the standard normal distribution is symmetric (about 0), the following is identity is true: 𝑃 𝑍≥𝑐 =𝑃(𝑍≤−𝑐). -1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587 0.1587 Because the standard normal distribution is symmetric (about 0), the following is identity is true: 𝑃 𝑍≥𝑐 =𝑃(𝑍≤−𝑐). -1.00

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! Key idea is to state the probability as 𝑃 𝑍≥𝑐 for some value positive 𝑐 so that we can use the table. 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587 𝑃 −1≤𝑍≤0.50 Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏) -1.00 0.50

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! Key idea is to state the probability as 𝑃 𝑍≥𝑐 for some value positive 𝑐 so that we can use the table. 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587 𝑃 −1≤𝑍≤0.50 =𝑃 𝑍≥−1 −𝑃 𝑍≥0.50 Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏) -1.00 0.50

Standard Normal Distribution Random variable 𝑍 has a standard normal distribution if 𝑍 is normally distributed 𝐸 𝑍 =𝜇=0 𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1 PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2 𝑃 𝑍≥𝑐 = 𝑐 ∞ 1 2𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? Still can’t do this by hand, so you have to use a table to look it up! Key idea is to state the probability as 𝑃 𝑍≥𝑐 for some value positive 𝑐 so that we can use the table. 𝑃 𝑍≥1 =0.1587 𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1 =1−0.1587=0.8413 𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587 𝑃 −1≤𝑍≤0.50 =𝑃 𝑍≥−1 −𝑃 𝑍≥0.50 =𝑃 𝑍≤1 −𝑃 𝑍≥0.50 = 1−𝑃 𝑍≥1 −𝑃 𝑍≥0.50 = 1−0.1587 −0.3085=0.8413−0.3085=0.5328 0.5328 Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏) -1.00 0.50

Standard Normal Distribution Suppose that we have the probability that something occurs p, and need to know the value of c beyond which the probability of occurrence is p. That is, 𝑃 𝑍≥𝑐 =𝑝 → 𝑐= ? If we assume that the probability distribution of Z follows a standard normal distribution, then this is simple to find using a table! 𝑝 c Now, let’s say we want to find the value of c, such that the probability of a value greater than c is equal to p.

Standard Normal Distribution Suppose the situation is like that given in the picture. What do we see? c is greater than 0. Then, we find in the table, the value of c that gives 𝑝≈0.3085 in the “Area beyond z in tail.” 0.3085 𝑐= 𝑧 𝑝 c 𝑃 𝑍≥𝑐 =0.3085 → 𝑐= 𝑧 0.3085 =0.50

Standard Normal Distribution Suppose the situation is like that given in the picture. What do we see? c is greater than 0. Then, we find in the table, the value of c that gives 𝑝≈0.3085 in the “Area beyond z in tail.” 0.3085 𝑐= 𝑧 𝑝 c = 0.50 𝑃 𝑍≥𝑐 =0.3085 → 𝑐= 𝑧 0.3085 =0.50

Standard Normal Distribution Suppose the situation is like that given in the picture. What do we see? c is greater than 0. We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝 Then, we first convert to finding the value of c in 𝑃 𝑍≥𝑐 =1−𝑝. that gives 1−𝑝≈0.1587 in the “Area beyond z in tail.” 0.8413 c 𝑃 𝑍≤𝑐 =1−𝑃 𝑍≥𝑐 =0.8413 → 𝑃 𝑍≥𝑐 =1−0.8413=0.1587 → 𝑐= 𝑧 0.1587 =1.00

Standard Normal Distribution Suppose the situation is like that given in the picture. What do we see? c is less than 0. We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝 Then, we first convert to finding c in 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =𝑝 that gives 𝑝≈0.1587 in the “Area beyond z in tail.” 0.1587 c 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =0.1587 → −𝑐= 𝑧 0.1587 =1.00→𝑐=−1.00

Standard Normal Distribution Suppose the situation is like that given in the picture. What do we see? c is less than 0. We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝 Then, we first convert to finding c in 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =𝑝 that gives 𝑝≈0.1587 in the “Area beyond z in tail.” 0.1587 c = -1.00 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =0.1587 → −𝑐= 𝑧 0.1587 =1.00→𝑐=−1.00