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

6 hours 48 min

step-by-step explanation:

68÷5=13.6

13.6×30=408min=6 hours 48 min.

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x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)