Question

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?

Answer 1

Answer: 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 :-

$P(X>44)=P(\dfrac{X-m}{s}>\dfrac{44-33}{9})\\\\=P(Z>\dfrac{11}{9})\\\\=P(Z>1.22)\\\\ =1-P(Z<1.22)\\\\=1-0.8888\\\\=0.1112$

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

Answer 2

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

• 34 is -0.5556 standard deviations from the mean.
• Using the normal distribution table,

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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