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A probability density function of a random variable is given by $f(x) = frac{1}{61} left(5 + frac{3}{sqrt{x}} ight)$ on the interval $[25, 36]$. Find the expected value, the variance, and the standard deviation. The expected value is $mu approx$ (Round to two decimal places as needed.) The variance is $Var(X) approx$ (Round the final answer to two decimal places as needed. Use the expected value rounded to two decimal places. Do not round any other intermediate values.) The standard deviation is $sigma approx$ (Round the final answer to two decimal places as needed. Use the expected value and variance rounded to two decimal places. Do not round any other intermediate values.)

          A probability density function of a random variable is given by $f(x) = frac{1}{61} left(5 + frac{3}{sqrt{x}}
ight)$ on the interval $[25, 36]$. Find the expected value, the variance, and the standard deviation.
The expected value is $mu approx$ 
(Round to two decimal places as needed.)
The variance is $Var(X) approx$
(Round the final answer to two decimal places as needed. Use the expected value rounded to two decimal places. Do not round any other intermediate values.)
The standard deviation is $sigma approx$
(Round the final answer to two decimal places as needed. Use the expected value and variance rounded to two decimal places. Do not round any other intermediate values.)

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$A probability density function of a random variable is given by f(x) = frac161 left(5 + frac3sqrtx ight) on the interval [25, 36]. Find the expected value, the variance, and the standard deviation. The expected value is mu approx (Round to two decimal places as needed.) The variance is Var(X) approx (Round the final answer to two decimal places as needed. Use the expected value rounded to two decimal places. Do not round any other intermediate values.) The standard deviation is sigma approx (Round the final answer to two decimal places as needed. Use the expected value and variance rounded to two decimal places. Do not round any other intermediate values.)$

Added by Mark G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Madhur L

06/23/2023

Step 1

For this, we need to satisfy the condition: $$\int_{25}^{36} c dx = 1$$ Integrating, we get: $$c(x) \Big|_{25}^{36} = c(36 - 25) = 11c = 1$$ Solving for c, we get: $$c = \frac{1}{11}$$ Now, the probability density function is: $$f(x) = \frac{1}{11}$$ Show more…

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A probability density function of a random variable is given by f(x) on the interval [25, 36]. Find the expected value, the variance, and the standard deviation. The expected value is E(X) ≈ (Round to two decimal places as needed): The variance Var(X) ≈ (Round the final answer to two decimal places as needed. Use the expected value rounded to two decimal places. Do not round any other intermediate values.) The standard deviation is σ ≈ (Round the final answer to two decimal places as needed. Use the expected value and variance rounded to two decimal places. Do not round any other intermediate values.)

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00:01 Hi, let's start the solution.

00:02 In this question, we have given that the probability density function of a random variable is given that is f of x is 1 upon 61 in bracket 5 plus 3 divided root x when 25 is less than equal to x less than equal to 36 and 0 otherwise.

00:26 Then the expected value of the random variable x is ex that is equal to integration minus infinity to infinity x f of x dx.

00:45 So that is equal to minus infinity to 25 x f of x dx plus 25 to 36 x f of x dx plus 36 to infinity x f of x dx.

01:09 So we get that is equal to 0 plus integration 25 to 36 5 x dx that is x into 61 x into 1 61 5 plus 3 divided by root x dx plus 0.

01:37 So we get that is equal to 1 upon 61 limit integration limit 25 to 36 5 x dx plus 3 upon 61 integration 25 to 36 x power 1 by 2 dx.

01:56 Now by integrating this here we get that is equal to 5 upon 61 integration of x is x square by 2 limit 25 to 36 plus 3 upon 61 integration of x power 1 by 2 is x power 3 by 2 divided by 3 by 2 limit 25 to 36.

02:22 Now by applying the limit here we get that is equal to 5 divided by 122 36 square minus 25 square plus 2 upon 61 in bracket we have 36 power 3 by 2 minus 25 power 3 by we get which is equal to 5 divided by 122 into 671 plus 2 divided by 61 into 91.

02:56 So we get that is equal to 3719 divided by 122.

03:03 Therefore we get the expected value is mu which is equal to 30 .48.

03:19 Now next we know that variance x equal to e x square minus e of x whole square...

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