The graph shows a probability distribution. P(X≤a)=0.5 What is the value of a?
2 answers:
Answer:
Step-by-step explanation:
We have been given an image of a normal distribution. We are asked to find the value of a for .
We need to find the value of X such that area under that value is less than or equal to 0.5.
We know that entire area under a normal curve is 1. Since the area under the required value is 0.5, which is half of the 1, so we need to find the center of our given distribution.
Upon looking at our given distribution, we can see that its center is at 2.
Since , therefore, the value of a is 2.
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Circle formula
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius
to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=
points (-3,2) and (1,5)
D=
D=
D=
D=
D=5
center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1
the point (-7,5) and (7,-1) lie on this circle
radius=5 units
the points (-7,5) and (-7,1) lie on this circle
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center
and r=radius
to find the radius
we are given one of the points and the center
distnace from them is the radius
distance formula
D=
points (-3,2) and (1,5)
D=
D=
D=
D=
D=5
center is -3,2
r=5
input
(x-(-3))^2+(y-2)^2=5^2
(x+3)^2+(y-2)^2=25 is equation
radius =5
input -7 for x and solve for y
(-7+3)^2+(y-2)^2=25
(-4)^2+(y-2)^2=25
16+(y-2)^2=25
minus 16
(y-2)^2=9
sqqrt
y-2=+/-3
add 2
y=2+/-3
y=5 or -1
the point (-7,5) and (7,-1) lie on this circle
radius=5 units
the points (-7,5) and (-7,1) lie on this circle