4. The weekly demand for propane gas (in 1000s of gallons)...

4. The weekly demand for propane gas (in 1000s of gallons) from a particular facility is a random variable X with pdf $f(x) \begin{cases} 2 \left(1 - \frac{1}{x^2} \right) & 1 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$ (a) What is the probability that more than 1500 gallons is demanded in a week? (2 points) (b) What is the CDF of X? (2 points) (c) Compute the expected value, variance, and standard deviation of X. (5 points) (d) Obtain an expression for the $p^{th}$ percentile of demand. What is the median demand? (3 points)

Transcript

00:01 So, the pdf is given by 2 into 1 minus 1 by x squared, 1 less than input x less than input 2 and 0 else.

00:12 This is the probability density function of the random variable.

00:14 Let's find the cdf first.

00:17 F of f of x, capital f of x is the cdf.

00:20 It is basically defined as negative infinity to x the pdf.

00:27 Case 1.

00:31 If suppose x is less than 1, then basically the cdf is.

00:35 Is 0 because an x is less than 1 the pdf is 0 because it is in the else part case 2 when x belongs to 1 to 2 now the cdf will be integration of 1 to x 2 into x 2 into 1 minus 1 by x squared d x so let's do the integration f of x is equal to integration of 1 minus 1 by x square is x plus 1 by x let's substitute the upper and lower limits it will be two times x plus 1 by x minus 2 this is the cdf when x belongs to 1 now case 3 when x is greater than or equals to greater than 2 then the cdf will be 1 because it will be the total area under the period which is 1 so finally the cdf of the random variable is given by 0 when x is less than 1 two times of x plus 1 by x minus 2 when x belongs to 1 to 2 and 1 x greater and 2 this is the cd now let us calculate the 100 pth percentile 100 pth percentile is basically is basically the p value is equal to negative infinity to alpha the pdf so basically it is f of alpha.

02:12 Obviously my f of alpha should be taken this because the hundred p to hundred p's percent higher the corresponding pay value should be equated to f of alpha with this branch.

02:22 So p is equal to 2 into alpha plus 1 by alpha minus 2.

02:26 So let's calculate alpha purely in terms of p you got a quadratic equation.

02:30 So alpha plus 1 by alpha is equal to p by 2 plus 2 or alpha square minus alpha into p by two plus two plus one is equal to zero which is a quadratic equation so let's all using the quadratic formula so alpha is equal to p by two plus two plus or minus we'll take plus because minus will not be feasible would of b square that is p by two plus two whole square minus four into one into one four divided by two so simplifying further you'll be you'll be getting p by four plus one plus half of under square root it is p square by four plus two p simplifying further so alpha is equal to one fourth of p plus four plus root of p square plus eight p this is the value of the hundred tp percent time now to find the population median new tilt just substitute p is equal to half because population median is basically 50th percent time fiftieth percentile is median and that is denoted mutil so substitute p is half so alpha is equal to one fourth of half plus four plus root of half square that is point to five plus eight into half that is four so if you just use calculator you'll be getting 1 .6403 so this is the answer expectation of the random variable is basically integration of 1 to 2 x times the pdf pdf is 2 into 1 minus 1 by x squared d x so let's find us so it is integration of 1 to 2 x, 2x minus 2x dx.

04:18 So let's integrate integration of 2 x is what? it is x squared.

04:23 X square minus 2 l n mod x.

04:27 So substitute with the limits.

04:29 So it will be 2 squares 4 minus 2 l n 2 and minus 1 and ln 1 is 0.

04:34 So it is 3 minus ln 4...

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