A normal population has a mean m=33 and standard deviation s = 9. What is the probability that a randomly chosen value will be greater than 44?

A normal population has a mean m=33 and standard deviation s = 9. the probability that a randomly chosen value will be greater than 44 is - 28.774%

Given:

mean m=33

standard deviation s = 9

probability = ?

Solution:

(34-39)/9 = -0.5556

which mean

P(Z<-0.5556) ≈ 0.28774

or 28.774%

Thus, A normal population has a mean m=33 and standard deviation s = 9. the probability that a randomly chosen value will be greater than 44 is - 28.774%

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0.1112

Step-by-step explanation:

Let X be a random variable .

Given: A normal population has a mean m=33 and standard deviation s = 9.

The probability that a randomly chosen value will be greater than 44 will be :-

Hence, the probability that a randomly chosen value will be greater than 44 = 0.1112

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

answer: it depends on what the tax percentage