6.1 Random Variables Random variable – variable whose value is a numerical outcome of a random phenomenon. Discrete – countable number of number of probable outcomes Continuous – measurable ex. Temperature, height, age
Discrete variables Discrete example Find P(x=0) Find P(x≤2) Cars owned x Number of families relative frequency or P(x) xP(x) 17 0.00425 1 877 0.21925 0.219 2 1767 0.44175 3 1107 0.27675 0.83 4 200 0.05 5 32 0.008 4000
Random variables Population mean μ also called “expected value” μ = Σ xP(x) variance σ2 = Σ (x- μ)2P(x) Standard deviation = σ Find μ and σ for last problem
Continuous variables What was the age distribution of nurses in Great Britain at the time of Florence Nightingale 1851 25466 nurses in Great Britain? Age Midpoint P(x) xP(x) x-μ (x-μ)2 (x-μ)2P(x) 20-29 24.5 0.057 1.4 -29.3 856 48.8 30-39 34.5 0.097 3.35 -19.3 371 36 40-49 44.5 0.197 8.77 -9.26 85.7 16.9 50-59 54.5 0.292 15.9 0.74 0.55 0.16 60-69 64.5 0.25 16.1 10.7 115 28.8 70-79 74.5 0.091 6.78 20.7 430 39.1 80+ 84.5 0.018 1.52 30.7 945 17 1.002 53.8 187 13.7
Continuous variables Draw a histogram of the distribution Find the probability that a nurse selected at random would be 60 or older. What is the expected age of nurses in 1851? Find the standard deviation.
Random variables USA Today reported that about 25% of all state prisoners released on Parole become repeat offenders while on parole. Suppose the parole board is examining 5 prisoners for parole. Let x= number of prisoners of the 5 on parole who become repeat offenders. Here is the probability distribution:
Random variables x 1 2 3 4 5 P(x) 0.24 0.4 0.26 0.09 0.02 1 2 3 4 5 P(x) 0.24 0.4 0.26 0.09 0.02 Find the probability that one or more of the 5 will be repeat offenders? Find the probability that 2 or more will be repeat offenders? Find the probability that 4 or more will be repeat offenders? Find the expected number of repeat offenders? Find the standard deviation.