16 B Notes

discrete random variable

-discrete random variables commonly arise from situations that involve counting something. -the probability model of a discrete random variable X assigns a probability between 0 and 1 to each possible value of X

continuous random variables

-infinite sample space -Situations that involve measuring something often results in a continuous random variable -continuous random variable x takes on all values in an interval of numbers. The probability distribution of X is described by a density curve. The probability of an event is the area under the density curve and above the values of x that make up the event -if generating number from 0 to 10, can't list out every possible value because there is an infinite value of x (in discrete model, only generating integers (like 0,1,2,3) and not all real numbers (like 1.555 or 2.34)) -a continuous random variable Y has infinitely many possible values. All continuous probability models assign probability 0 to every individual outcome. Only intervals of values have positive probability

normal probability distributions continuous model -the heights of young women closely follow the normal distribution with mean µ=64 inches and standard deviation σ=2.7 inches. Now choose one young woman at random. Call her height Y. If we repeat the random choice very many times, the distribution of values of Y is the same normal distribution that describes the heights of all young women. what's the probability that the chosen woman is between 68 and 70 inches tall? what about probability that find woman's height that is between 68 and 70 given that y is less than 70?

-normal distribution = continuous model because x has indefinite sample space P(68≤x≤70) = normal cdf (68, 70, 2.7, 64) = 5.62% -64 is mean, middle of normal distribution normalcdf(lower, upper, st dev, mean) P(68≤x≤70/ x≤70) = P(both)/P(x≤70) = 5.62%/normalcdf (- infinity, 70, 64, 3.7) = 5.62%/98.68% = 5.69% -negative infinity on calc is -1E + 99

consider the piecewise defined function: f(x):{ 1/4, if 0 < x < 1 (5-x)/16, if 1 < x < 5 x-5/2, if 5 ≤ x < 6 0, otherwise Plot this probability density function and compute the 50th percentile of X

-number generator from 0 to 6 [0, 6] -f(x) is the probability -x axis is 0 to 6 -plot graph: from 0 to 1, horizontal line at 1/4 on y axis; from 1 to 5, line going down with slope of -1/16 until x=5 where y=0; then from 5 to 6, the slope goes back up (bc positive slope); and then after 6, nothing happens - 50th percentile is the median (50th percentile is the % of values that are 50 or lower) 1. figure out where x should be so that the area to the left is 50% 2. area from 0 to 1 where the rectangle is, is 25% (because bh= (1)(1/4)) 3. have to find where x is after the rectangle to have another 25% 4. 1/4=1/2(x-1)(1/4 + (5-x/16)), area of trapezoid right next to the rectangle -it is x-1 because you are not including that 0 to 1 in the beginning but only finding the height of the trapezoid where it is 1 to x so have to subtract that 0 to 1 part -to find b2, plug x into the equation 4. 5-x = +/- rad 8 -x = -5 +/- rad 8 x= 5 + rad 8 or 5 - rad 8 5. x= 5 - rad = 2.171 because 5 + rad 8 is greater than 6 (this is where x needs to be so that 50% of area is to the left) OR 1. you can use the two triangles from 1 to 6 because of the two slope lines 2. the area of the triangle from 5 to 6 is 1/4 because 1/2bh= 1/2(1/2)(1) = 1/4 3. so, the other triangle has to be from x to 5 and equal to 1/4 4. 1/4=1/2(5-x)((5-x)/16) 1/2= (5-x)^2/16 8=(5-x)^2 5. get the same answer, it is 5-x because you are not including that 0 to x in the beginning but only finding where it is x to 5 so have to subtract that x from 5 to account for the 0 to x part (it would only be x-5 if x > 5 and you need to find 5 to x and not include the 0 to 5 part)

consider the piecewise defined function: f(x):{ x+1/2 0≤ x ≤ a 0 otherwise Find value of a to make this a legitimate probability model (b) P(x < 1/2 | x < 3/4)

-random number generator from [0,a] (generating real numbers not integers) so generating continuous models -f(x) is the probability (y-axis) -x-axis is 0 to a -P(sample space) must equal 100% in order to have a legitimate probability model -from 0 to a, we have a line where the intercept is at 1/2 -beyond a, there is nothing because f(x)=0 other than 0 to a -density curve is area below the line which is the probability -so, have to make sure area under density curve has to be 100% Trapezoid area: 1. 1/2h(b1 + b2) 2. when you plug in a into the equation to find what b2 is on the y-axis, then you get a + 1/2 3. b1=1/2 because if you plug 0 into the equation to find b1 on the y-axis then it is 1/2 4. 1=1/2(a)(1/2+ a + 1/2) 5. 1=1/2a(a+1) 4. a^2 + a -2 = 0 5. (a+2)(a-1)=0 6. a=1 because a can not be less than 0 (b) 1. P(x < 1/2 n x < 3/4) / P(x < 3/4) 2. find trapezoid area of P(x<1/2), which is 1/2(1/2)(1/2+1) = 3/8 (you get b2 by plugging 1/2 into the equation to find y) 3. then find trapezoid area of P(x < 3/4) which is 1/2(3/4)(1/2+5/4)= 21/32 4. P(x < 1/2 n x < 3/4) is the same as P(x < 1/2) so P(x < 1/2 n x < 3/4)=3/8 5. 3/8 / 21/32 = 4/7

The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and standard deviation 7. Volunteers in the top 20 percent of hours worked will receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given the number of hours the volunteer worked in less than 90?

1. normal distribution, normal density curve so you can use normalcdf 2. conditional probability with "given" so it has to be P(x=top 20% n x ≤ 90 hours worked)/P(x ≤ 90 hours) 3. find P(x=top 20%) so you do invnorm(0.8, 7, 80) on graphing calculator to find out where x is on the density curve to have the top 20% 4. then find where P(x=top 20% n x ≤ 90 hours worked) which is normalcdf(85.89, 90, 7, 80) 5. then, to find the denominator it is normalcdf(-infinity, 90, 7, 80) 6. then divide those two numbers to get your answer

calculator function for normalcdf and invcdf

click on "OPTN" then click on "STAT" then click on "DIST" then click on "NORM" then... click on "Ncd" and input normCD(lowest, highest, st. deviation, mean) -infinity= -1E + 99 +infinity= 1E + 99 or if you want to find where x is on the normal density curve given a desired probability (ex: top 20% or bottom 40%) -gives percentiles, like 99th percentile is top 1%; 5th percentile is bottom 5% use "INVNORM" and input InvNormCD(left area, st. deviation, mean) -left area is like you want to find where x is top 20% so you use the left area to find it which is 80% (BUT WRITE IN DECIMAL FORM)