1. 1 Continuous random variables f(x) x

2. 2 Continuous random variables A discrete random variable has values that are isolated numbers, e.g.: Number of boys in a family number of heads in 10 flips of a coin A continuous random variable has values over an entire interval, e.g.: Height of people All values in the interval [2.2,8.2]

3. 3 Difference between discrete and continuous rv’s In the random numbers table, every digit 0,1,…,9 has the same probability of 0.1 to be selected. S={0,1,2,3,4,5,6,7,8,9} Probability histogram: 0 1 2 3 4 5 6 7 8 9

4. 4 Difference between discrete and continuous rv’s Now suppose that we want to choose at random a number in [0,1]. You can visualize such a random number by thinking of a spinner that turns freely on its axis and slowly comes to stop: ½ ¼ 0 ¾ In this case, the sample space is an interval: S={all numbers x such that 0≤x≤1}

5. 5 Difference between discrete and continuous rv’s S={all numbers x such that 0≤x≤1} We want all possible outcomes to be equally likely. However Impossible to assign a probability to each x because there are infinitely many x’s. ½ ¼ 0 ¾

6. 6 Difference between discrete and continuous rv’s Instead, we assign probabilities to intervals under a density curve. For the spinner example we obtain the following curve: Area under the curve=1 0 1 Height=1

7. 7 Difference between discrete and continuous rv’s What is the probability of a number between 0.2 and 0.6? 0.2.6 1 Area = (.6-.2)1=0.4 =p(.2≤x≤.6) p(X≥7)= Height=1.3

8. 8 Continuous random variable 1.Takes all values in an interval 2.The probability distribution is described by a density curve f(x) 3.f(x)≥0 for all x 4.The probability of any event is the area under the density curve and above the values of X that make up the event

9. 9 Continuous random variable 5. All continuous distributions assign probability zero to every individual outcome 0.2 1 P(X=.2)=0 Since P(X=.2)=0  P(X>.2)=p(X≥.2) (this is true only for continuous rv’s)

10. 10 Only intervals of values have positive probability: 0 1.79.81 P(.79≤X≤.81)=0.02 0 1.799.801.7999.8001 P(.799≤X≤.801)=0.002 P(.7999≤X≤.8001)=0.0002

11. 11 An interval of size zero has a probability of zero: 0 1.80 P(.8≤X≤.8)= P(X=8)=0

12. 12 Example Let X be a random number between 0 and 1 with uniform density curve in the interval [0,1] 0 1 f(x) 1 P(0≤X≤0.4)= P(0.4≤X≤1)= P(X=0.5)= P(0.3≤X≤0.5)= P(0.3<X<0.5)= P(0.226≤X≤0.713)=.4.6 0 0.2 p(X≤0.713)-p(X≤.226)=.713-.226=.487

13. 13 Example A random number generator produces numbers from 1 to 3 with a uniform density curve as follows. 1 3 f(x) 1.What is the height of the density curve? Since the area under the curve is 1, the height must be ½ 2. P(1≤X≤3)= P(2≤X≤3)= P(1≤X≤1.8)= P(1.8<X≤2.5)= 1 p(X≤3)-p(X≤2)=1-0.5 p(X≤1.8)-p(X≤1)=(0.8)(½)-0=.4 p(X≤2.5)-p(X<1.8)=(2.5-1)(½)-(1.8-1)(½)=.75-.4=.35

14. 14 Example For of the functions sketched in (a)–(d) state whether it could/could not be the probability density function of a continuous random variable? 0 1 2 1 0.5 0 1 2 1 1 1 (a) (b) (c) (d) f(x)

15. 15 Example For the following probability density function, which of the two intervals is assigned a higher probability: P(0<X<.5) Or P(1.5<X<2) ? 1 0 1 2 f(x)

16. 16 Example A continuous random variable may have various forms: for example: f(x) x x x x

17. 17 Normal distribution as a continuous distribution N(μ,σ) is a normal distribution with mean μ and standard deviation σ If X~N(μ,σ), then μ Density curve

18. 18 Normal distribution as a continuous distribution Example Scores in a certain exam are distributed normally with Mean 80 and SD 12. 1. What proportion of students receive a score higher than 86? P(X>86)= = p(Z>.5) = 1- Ф (.5) = 1-.6915=.3085  30.85% of the students received a score higher than 86

19. 19 2. What is the probability to receive a score that differs from the mean by no more than 10 points? p(70≤X≤90)= = =p(-.833≤Z≤.833)= Ф (.8333)- Ф (-.8333)=.7967-.2033=.5936 807090

20. 20 3. Find the third quarter of the scores (find Q 3 ) We first find Q3 on the z-scale and then convert it to x-scale. z 0.75 =0.675 (  p(Z≤.675)=0.75 ) Q 3 = 0.675(12)+80 = 88.1 80 Q 3

21. 21 4. The lowest 5% of the scores are below which score? We first find x on the z-scale and then convert it to x-scale. z 0.05 =-1.645 (  p(Z≤-1.645)=0.05 )  x=-1.6-1.645(12)+80 = 60.26 x 80 5%