320 likes | 549 Views
Probability Distributions. Random Variable A numerical outcome of a random experiment Can be discrete or continuous Generically, x Probability Distribution The pattern of probabilities associated with all of the random variables for a specific experiment Can be a table, formula, or graph
E N D
Probability Distributions • Random Variable • A numerical outcome of a random experiment • Can be discrete or continuous • Generically, x • Probability Distribution • The pattern of probabilities associated with all of the random variables for a specific experiment • Can be a table, formula, or graph • Generically, f(x) • Examples • Binomial (but won’t cover here) • Uniform • Normal or bell-shaped distribution
Birth of a Distribution Class Width = 10 Cyberland Wages
Birth of a Distribution Class Width = 5
Birth of a Distribution Class Width = 2
Birth of a Distribution Class Width = 1
Birth of a Distribution Class Width = Very Small
Uniform Distribution f(x) Area = 1 1 / (b-a) x a b
Normal Distribution Bell-shaped, symmetrical distribution f(x) x
Normal Distributions =1 =2 =3 = 5 12 -2
Normal Distributions Same ,Different
Normal Distributions 68.26% - +
Normal Distributions 95.44% -2 +2
Normal Distributions 99.72% -3 +3
Standard Normal Distribution z = 0 z = 1 If x has a normal distribution… z 0
t Distribution Looks like a normal distribution, Specific thickness depends on degrees of freedom but has thicker tails -3.5 3.5 0
t Distribution 5 d.f. 10 d.f. 30 d.f. 100 d.f. d.f (normal) Specific thickness depends on degrees of freedom -3.5 3.5 0
Find the Probabilities • P(z > 2.36) • P(t > -1.02) with 5 degrees of freedom • P(-0.95 < z < 1.93) • P(-0.95 < t < -0.07) with 100 degrees of freedom • Find z* such that P(z < z*) = 0.719 • Find z0.025 such that P(z > z0.025) = 0.025 • Find t0.025 such that P(t > t0.025) = 0.025 with 5 degrees of freedom
z/2 Standard Normal Distribution (z) P(z < -z/2)) = /2 P(z > z/2) = /2 P(-z/2 < z < z/2)= 1 - /2 -z/2 0 z/2
z/2 for =0.05 Standard Normal Distribution (z) P(z < ) = 0.025 P(z > ) = 0.025 P( < z < )= 0.95 -z0.025 0 z0.025 ? ?
t/2 for =0.05, df=5 t distribution with 5 degrees of freedom P(t < ) = 0.025 P(t > ) = 0.025 P( < t < )= 0.95 -t0.025 0 t0.025 ? ?
2 Distribution Specific skewness depends on degrees of freedom 0
2 Distribution Specific skewness depends on degrees of freedom 5 d.f 10 d.f 15 d.f 0
2 Distribution 10 d.f P(2 > 18.307) = 0.05 P(2 < 18.307) = 0.95 0 18.307
F Distribution Specific skewness depends on a pair of degrees of freedom (df1, df2) 0
F Distribution P(F < 3.02) = 0.95 9 and 10 d.f P(F > 3.02) = 0.05 0 3.02
Probability Distributions Normal & t 2 Different shapes and df’s, but SAME LOGIC ! F
In Excel • To find probability above a value x • =1-NORMSDIST(x) • =TDIST(x,df,1) [1=1-tail] • =CHIDIST(x,df) • =FDIST(x,df1,df2) • To find value with p% above (e.g., 0.05) • =NORMSINV(p) • =TINV(p,df) • =CHIINV(p,df) • =FINV(p,df1,df2)
Word Problem • From past experience, the management of a well-known fast food restaurant estimates that the number of weekly customers at a particular location is normally distributed, with a mean of 5000 and a standard deviation of 800 customers. • What is the probability that on a given week the number of customers will be between 4760 and 5800? • What is the probability of a week with more than 6500 customers? • For 90% of the weeks, the number of customers should exceed what amount?