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%

-2

7 - 9 = -2

reversed:

-2 + 9 = 7

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560