If y is normal with μ = 200 and σ = 12, what is the probability that y will be between 186 and 210?

Normalizing the bounds,

y=186, z = (186-200)/12 = -1.17

y = 210, z = (210 - 200)/12 = 0.83

We want the area of the standard normal between those. Looking it up in the cumulative table,

Ф(1.17) = 0.87900

Ф(0.83) = 0.79673

We need Ф(-1.17) which is 1 - Ф(1.17)

Then the probability we seek is

p = Ф(0.83) - Ф(-1.17) = Ф(0.83) + Ф(1.17) - 1 = 0.87900 + 0.79673 - 1 = 0.67573

67.6%

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

1,589.63 in.^3

step-by-step explanation:

i hope this you : d