Respuesta :
Answer:
a) [tex] f(x) = \frac{1}{b-a}= \frac{1}{17-(-8)}= \frac{1}{25}=0.04, -8 \leq X \leq 17[/tex]
And the height for the density function would be 0.04
b) [tex] Var(X) =\frac{(b-a)^2}{12}= \frac{(17+8)^2}{12} = 52.08333[/tex]
And the deviation is given by:
[tex] Sd(X) = \sqrt{52.08333}= 7.2169[/tex]
c) [tex] P(X \leq -5) = F(-5) = \frac{-5+8}{17+8}= 0.12[/tex]
Step-by-step explanation:
For this case we assume the X is the random variable of interest and is given by:
[tex] X \sim Unif (a=-8, b=17)[/tex]
Part a
The density function is given by:
[tex] f(x) = \frac{1}{b-a}= \frac{1}{17-(-8)}= \frac{1}{25}=0.04, -8 \leq X \leq 17[/tex]
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:
[tex] Var(X) =\frac{(b-a)^2}{12}= \frac{(17+8)^2}{12} = 52.08333[/tex]
And the deviation is given by:
[tex] Sd(X) = \sqrt{52.08333}= 7.2169[/tex]
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:
[tex] F(X) = \frac{x-a}{b-a}[/tex]
And we got:
[tex] P(X \leq -5) = F(-5) = \frac{-5+8}{17+8}= 0.12[/tex]
