Suppose a random variable X is best described by a uniform probability distribution with range 1 to 5. Find the value of that makes the following probability statements true. a) P(X <-a)= 0.95
b) P(X c) P(X d) P(X ->a)= 0.89
e) P(X >a)= 0.31

a) 4.8

b) 2.96

c) 4.4

d) 1.44

e) 3.76

Step-by-step explanation:

What we will do is solve point by point, knowing the following:

Fx (x) = P (X <= x) = (x - 1) / 4

a) P (X <-a) = 0.95

Fx (a) = 0.95

(a -1) / 4 = 0.95

a = 1 + 0.95 * 4

a = 4.8

b) P (X <a) = 0.49

Fx (a) = 0.49

(a -1) / 4 = 0.49

a = 1 + 0.49 * 4

a = 2.96

c) P (X <a) = 0.85

Fx (a) = 0.85

(a -1) / 4 = 0.55

a = 1 + 0.85 * 4

a = 4.4

d) P (X> a) = 0.89

P (X <a) = 1 - 0.89 = 0.11

Fx (a) = 0.11

(a -1) / 4 = 0.11

a = 1 + 0.11 * 4

a = 1.44

e) P (X> a) = 0.31

P (X <a) = 1 - 0.31 = 0.69

Fx (a) = 0.69

(a -1) / 4 = 0.69

a = 1 + 0.69 * 4

a = 3.76

the answer would be the last one (constat) she stays at the same rat every two mintues it goes up by 0.4

step-by-step explanation:

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this