A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750. a. Calculate the mean and the stan

Question

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750. a. Calculate the mean and the standard deviation for the distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.) b. What is the probability that X is less than 730

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Vivian 2 months 2021-10-13T23:30:06+00:00 1 Answer 0 views 0

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    2021-10-13T23:31:37+00:00

    Answer:

    a)  E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710

    The variance is given by:

     Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333

    And the deviation is just the square root of the variance and we got:

     Sd(X) = \sqrt{5333.333}= 23.09

    b)  P(X<730)

    And for this case we can use the cumulative distribution given by:

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

    And replacing we got:

     P(X<730)= F(730) = \frac{730-670}{750-670}=0.75

    Step-by-step explanation:

    For this case we assume that X is our random variable and we know that the distribution for X is given by:

     X \sim Unif(a=670, b = 750)

    Part a

    For this case the expected value is given by:

     E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710

    The variance is given by:

     Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333

    And the deviation is just the square root of the variance and we got:

     Sd(X) = \sqrt{5333.333}= 23.09

    Part b

    For this case we want to find this probability:

     P(X<730)

    And for this case we can use the cumulative distribution given by:

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

    And replacing we got:

     P(X<730)= F(730) = \frac{730-670}{750-670}=0.75

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