Continuous Probability Distributions

Chapter Continuous Probability Distributions 7.1 Probability Density Function A continuous random variable has an uncountable infinite number of values in the interval (a,b) To calculate probabilities of continuous random variables we define a probability density function f(x), which satisfies the following conditions Area = – f(x) is non-negative, – The total area under the curve representing f(x) is equal to The probability that a continuous variable X will assume any particular value is zero The probability that x falls between ‘a’ and ‘b’ is the area under the graph of f(x) between ‘a’ and ‘b’ P(a≤ x≤ b) a b Uniform Distribution A random variable X is said to be uniformly distributed if its density function is f ( x) = with b−a a+b E(X) = f(X) a b a ≤ x ≤ b (b − a) V(X ) = 12 X Example 7.1: The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons Find the probability that sales are: Between 2,500 and 3,000 gallons P(2500≤ X≤ 3000) = (3000-2500)(1/3000) = 1667 1/3000 2000 2500 3000 5000 x 7.2 Normal Distribution A random variable X with mean µ and variance σ is normally distributed if its probability density function is We denote a normal distribution by N(µ, σ ) Normal distributions range from minus infinity to plus infinity x − µ   x − µ   −−(1(1/ /22)) σ   σ  e 1 ∞≤≤ xx≤≤∞ ∞ ff((xx))== e −−∞ σσ 22ππ where ππ==33 14159 14159 and and ee ==22 71828 71828 where The Shape of the Normal Distribution µ bell shaped, and symmetrical around µ Increasing the mean shifts the curve to the right… 8.8 Increasing the standard deviation “flattens” the curve… 8.9 Two facts help calculate normal probabilities: - The normal distribution is symmetrical Any random variable (rv) X having N(µ , σ ) can be transformed into a rv Z having N(0 , ) , called the “STANDARD NORMAL DISTRIBUTION” or the Z-distribution by X −− µµ X ZZ == σσ 10 Checking the required conditions: The Student t distribution is robust, which means that if the population is non-normal, the results of the t-test and confidence interval estimate are still valid provided that the population is “not extremely non-normal” 14 12 10 30 25 20 15 10 400 425 450 475 500 525 550 575 More -4 14 22 30 108 More Example Assume the content of a can of Bubbly cola is Normally distributed Design a test to see whether its manufacturer adequately fills their 1-liter (i.e., 1000 mls) bottles H0 : μ = 1, 000mls HA: μ < 1, 000mls (1-side test) (adequately) (inadequately) α = (This might be just a class-room project) n = 25 (small sample size) -> Use the t 24 -distribution Reject H0 at α = 10 Reject region: T < −1.32 −1.32 Critical value 109 11.2 Inference About a Population Variance This statistic is (n − 1) s σ If the population is normally distributed, it has a 22 χχnn−−11 Chi-squared distribution, with df = n-1: The Chi-squared distribution 110 Testing the population variance – Left hand tail test Example 3: A container-filling machine is considered to fill liter containers consistently if the variance of the filling is less than cc (.001 liter) A random sample of 25 1-liter fills was taken, and s2=.6333 Do these data support the belief that the variance is less than 1cc at 5% significance level? 111 2 Step 1:H0:σ = 1; H1:σ < Step 2: α = 0.05 Step 3: n= 25, use Step 4: Reject Region χχ 22 nn−−11 == χχ 22 24 24 χχ 22 11−−αα, ,nn−−11 Step 5: == χχ 22 95, ,24 24 95 13 85 85 ==13 Χ224 95 ( n − ) s χ = 13.85 σ Critical value (25 − 10)(.6333) = = 15.20 Step 6: Do not reject the null hypothesis 112 Testing the population variance – Right hand tail test; Two tail test; • A right hand tail test: – – • A two tail test – H0: σ = value H1: σ > value – H0: σ = value H1: σ ≠ value Rejection region: Rejection region χχ ≥≥χχ 22 22 αα, n, n−−11 χχ ≤≤χχ 22 22 11−−αα22, n, n−−11 oror χχ ≥≥χχ 22 22 αα22, ,nn−−11 113 From the following probability statement Estimating the population variance P(χ 21-α /2 < χ < χ 2α /2) = 1-α we have (by substituting χ = [(n - 1)s2]/σ 2.) 2 (n − 1)s 2 (n − 1)s (n − 1)s (n − 1) s < σ < < σ < 22 22 χ χ χ αα//22 χ11−−αα//22 This is the confidence interval for σ with 1-α % confidence level 114 Example 4: Estimate the variance of fills in example with 99% confidence (n − 1) s ( n − ) s < σ < χα / χ12−α / ? ? and n(1-p) > 5], pˆ is approximately normally distributed, with 117 µ = p and σ = p(1 - p)/n Test statistic for p Interval estimator for p (1-α confidence level) pˆ−−pp ˆ p ZZ == pp((11−−pp))//nn where np np>>55 and and nn((11−−pp))>>55 where pˆpˆ±± zzαα/ /22 pˆpˆ((11−−pˆpˆ))//nn provided nnpˆpˆ >> 55 and and nn((11−−pˆpˆ)) >> 55 provided 118 Example 11.5 (Predicting the winner in election day): Voters are asked by a certain network to participate in an exit poll in order to predict the winner on election day Based on the data presented in Xm12.5.xls (where 1=Democrat, and 2=Republican), can the network conclude that the republican candidate will win the state college vote? 119 Step 1: H0: p = 5; H1: p > Step 2: α = 0.05 Step 3: n= 765, use the Z-dist Step 4: Reject Region Step 5: z=  p− p p (1 − p ) / n 532 − = = 1.77 5(1 − 5) / 765 Step 6: Reject the null hypothesis Z 1.645 Critical value 120 Example (marketing In a survey of 2000 Estimating theapplication): Proportion TV viewers at 11.40 p.m on a certain night, 226 indicated they watched “The Tonight Show” Estimate the number of TVs tuned to the Tonight Show in a typical night, if there are 100 million potential television sets Use 95% confidence level pˆ ± zα / pˆ (1 − pˆ ) / n = 113 ± 1.96 113(.887) / 2000 = 113 ± 014 226/2000 = 113 1-.113 = 887 121 FlowchartDescribe of Techniques a Population Data Type? Interval Nominal Type of descriptive measurement? Central Location t test & estimator of u z test & estimator of p Variability X2 Χ2 test & estimator of d 12.122 ...7.1 Probability Density Function A continuous random variable has an uncountable infinite number of values in the interval (a,b) To calculate probabilities of continuous random... define a probability density function f(x), which satisfies the following conditions Area = – f(x) is non-negative, – The total area under the curve representing f(x) is equal to The probability. .. the curve representing f(x) is equal to The probability that a continuous variable X will assume any particular value is zero The probability that x falls between ‘a’ and ‘b’ is the area under