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

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step-by-step explanation:

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true