Respuesta :
Answer:
a. Height=0.0555
b. Mean=7
SD=4.0415
c. P(x≤1)=0.1665
Step-by-step explanation:
a. If we have a continous uniform distribution, we have a density function that is a constant:
[tex]f(x)=C[/tex]
In the other hand, the cumulative density function across the entire region should be one:
[tex]F(x)=\int\limits^{-2}_{16} {f(x)} \, dx =\int\limits^{-2}_{16} {C} \, dx=1\\\\C(16-(-2))=1\\\\C*18=1\\\\C=1/18=0.0555[/tex]
Then the height C of the density function is 0.0555.
b. The mean of this distribution is:
[tex]M=\frac{1}{2}(a+b)=\frac{1}{2}(-2+16) =\frac{1}{2}* 14=7[/tex]
The standard deviation of this distribution is:
[tex]S.D.=\frac{a+b}{\sqrt{12}} =\frac{-2+16}{\sqrt{12}} =4.0415[/tex]
c.
[tex]P(x\leq1)=0.0555*\int\limits^1_{-2} {} \, dx =0.0555*(1-(-2))=0.0555*3=0.1665[/tex]
Using the uniform distribution, it is found that:
a) The height is of [tex]\frac{1}{18}[/tex].
b) The mean is of 9 and the standard deviation is of 5.2.
c) P(X ≤ 1) = 0.1667.
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
In this problem, the bounds are [tex]a = -2, b = 16[/tex].
Item a:
The height of an uniform distribution is:
[tex]\frac{1}{b - a}[/tex]
In this problem:
[tex]\frac{1}{16 - (-2)} = \frac{1}{18}[/tex]
The height is of [tex]\frac{1}{18}[/tex].
Item b:
The mean is:
[tex]M = \frac{b - a}{2} = \frac{16 - (-2)}{2} = 9[/tex]
The standard deviation is:
[tex]S = \sqrt{\frac{(b-a)^2}{12}} = \sqrt{\frac{18^2}{12}} = 5.2[/tex]
The mean is of 9 and the standard deviation is of 5.2.
Item c:
[tex]P(X < 1) = \frac{1 - (-2)}{16 - (-2)} = 0.1667[/tex]
Then
P(X ≤ 1) = 0.1667.
A similar problem is given at https://brainly.com/question/14683948
