A random variable X follows the continuous uniform distribution with a lower bound of −8 and an upper bound of 17. a. What is the height of

Question

A random variable X follows the continuous uniform distribution with a lower bound of −8 and an upper bound of 17. a. What is the height of the density function f(x)?

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Delilah 2 weeks 2021-10-11T11:34:31+00:00 1 Answer 0

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    2021-10-11T11:35:47+00:00

    Answer:

    a)  f(x) = \frac{1}{b-a}= \frac{1}{17-(-8)}= \frac{1}{25}=0.04, -8 \leq X \leq 17

    And the height for the density function would be 0.04

    b)  Var(X) =\frac{(b-a)^2}{12}= \frac{(17+8)^2}{12} = 52.08333

    And the deviation is given by:

     Sd(X) = \sqrt{52.08333}= 7.2169

    c)  P(X \leq -5) = F(-5) = \frac{-5+8}{17+8}= 0.12

    Step-by-step explanation:

    For this case we assume the X is the random variable of interest and is given by:

     X \sim Unif (a=-8, b=17)

    Part a

    The density function is given by:

     f(x) = \frac{1}{b-a}= \frac{1}{17-(-8)}= \frac{1}{25}=0.04, -8 \leq X \leq 17

    And the height for the density function would be 0.04

    Part b What are the mean and the standard deviation for the distribution? (Round your answers to 2 decimal places.)

    For this case the variance is given by:

     Var(X) =\frac{(b-a)^2}{12}= \frac{(17+8)^2}{12} = 52.08333

    And the deviation is given by:

     Sd(X) = \sqrt{52.08333}= 7.2169

    Part c: Calculate P(X ≤ −5). (Round intermediate calculations to 4 decimal places and final answer to 4 decimal places.)


    For this case we can use the cumulative distribution function given by:

     F(X) = \frac{x-a}{b-a}

    And we got:

     P(X \leq -5) = F(-5) = \frac{-5+8}{17+8}= 0.12

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