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The reaction time (in seconds) to a stimulus is a continuous random variable with pdf$$f(x)=\left\{\begin{array}{cc}{\frac{3}{2 x^{2}}} & {1 \leq x \leq 3} \\ {0} & {\text { otherwise }}\end{array}\right.$$(a) Obtain the cdf.(b) What is the probability that reaction time is at most 2.5 s? Between 1.5 and 2.5 s?(c) Compute the expected reaction time.(d) Compute the standard deviation of reaction time.(e) If an individual takes more than 1.5 s to react, a light comes on and stays on either until one further second has elapsed or until the person reacts (whichever happens first). Determine the expected amount of time that the light remains lit. [Hint: Let $h(X)=$ the time that the light is on as a function of reaction time $X . ]$

(a) F(x) = 1.5(1 1/x) for 1 ? x 3, = 0 for x < 1, = 1 for x > 3 (b) .9, .4 (c) 1.648 s (d) .553 s (e) .267 s

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Temple University

University of North Carolina at Chapel Hill

Piedmont College

Idaho State University

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the reaction time to a certain stimulus is defined with the following probability density function and for part A were asked to find the CDF of the function. To find the CDF, we can integrate over the domain where the PdF is greater than zero. So that's from 1 to 3, and this comes out to 3/2 times one minus one over X For part B. We want to probabilities. So the first is the probability that the reaction time is at most 2.5 seconds. So this is equal to the cumulative probability at 2.5, which is the CDF at 2.5, and that comes out to 0.9. So there's a 90% chance that the reaction time is less than or equal to 2.5 seconds. The other probability is the probability that your reaction time is between 1.5 and 2.5 seconds. This is equal to the cumulative probability at 2.5, minus the cumulative probability at 1.5. This comes out 2.9 minus 0.5, which is 0.4 now for Part C. We're as to calculate the expected reaction time. That's E X. So that is the integral over the non zero domain of the pdf of X Times the PdF. So it's 3/2 times the natural log rhythm of X evaluated over the interval 123 and this comes out to 1.648 for Part D. We want to calculate the standard deviation now. The way we'll do this is we'll use the variance shortcut formula, which says that the variance is equal to the expectation on X squared, minus the expected value squared. So to use this, we will first find the expected expected value of X squared, and this comes out to three. And so we have the variance of X being equal to three squared. Or rather, that's just three. Eat the expected value of X squared minus the expected value squared. So that's 1.648 squared and this comes out 2.284 If we take the square root of that, we get the standard deviation, and that is 0.553 for Part E. We're told that if an individual takes more than 1.5 seconds to react. Ah, light comes on and it stays on for either one second or until the person reacts. Whichever duration is shorter, and we're as to find the expected length of time that the light stays on for were given a hint in the question to write the time that the light stays on as a function of the reaction time so we can call that h of X So X is the reaction time h of X. Now, if the reaction time is less than 1.5 seconds, then the duration that the light is on is zero between 1.5 and 2.5 seconds. The duration that the light is on is the reaction time, minus the 1.5 seconds that must elapse before the light turns on. And then, for any reaction time that is greater than 2.5 seconds, Then the light will just have been on for the one second turns off automatically after one second if the person still hasn't reacted. So in that case, H of X is one. So now we want to find the expected duration that the light is on. That is the expected value of HDX. So for X lesson 1.50 So that term will be zero for X from 1.5 2, 2.5. We have X minus 1.5 times the pdf of X. We're making use of this fact now for X is greater than 2.5. So this should be a three, not a 3.5 X is only defined to be none. Zero up to three. So we evaluate this from 1.5 to 2.5 and this wasn't this one is evaluated from 2.5 to 3 and we get zero point 166 plus 0.1, which gives us 0.266 So the light is expected to be on for just over a quarter of a second.

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