Assuming that the heights of college women are normally distributed with mean 65 inches and standard deviation 3.5 inches, what percentage of women are shorter than 68.5 inches

the percentage of women that are shorter than 68.5 inches is 84.1%

Step-by-step explanation:

Given that;

mean μ = 65

standard deviation σ = 3.5

percentage of women are shorter than 68.5 inches = ?

P( x < 68.5 ) = P(( x-μ)/σ)

we substitute

⇒ ( 68.5-65)/3.5)

⇒ P(z < 1)

form standard normal table

P( x < 68.5 ) = P(z < 1) = 0.84134 ≈ 84.1%

Therefore, the percentage of women that are shorter than 68.5 inches is 84.1%

do i do it on a graph?

step-by-step explanation:

(3x-8) or (3x+8)

step-by-step explanation: