Respuesta :
Answer:
Step-by-step explanation:
We would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = randomly chosen values.
µ = mean
σ = standard deviation
From the information given,
µ = 9
σ = 5
1) The proportion of the population that is less than 20 is expressed as
P(x < 20)
For x = 20
z = (20 - 9)/5 = 2.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.986
P(x < 20) = 0.986
2) The probability that a randomly chosen value will be greater than 6 is expressed as
P(x > 6) = 1 - P(x ≤ 6)
For x = 6
z = (6 - 9)/5 = - 0.6
Looking at the normal distribution table, the probability corresponding to the z score is 0.27
P(x > 6) = 1 - 0.27 = 0.73
Z-score helps us to understand how far is the data from the mean. The probability that a randomly chosen value will be greater than 6 is 0.2743.
What is Z-score?
A Z-score helps us to understand how far is the data from the mean. It is a measure of how many times the data is above or below the mean. It is given by the formula,
[tex]Z = \dfrac{X- \mu}{\sigma}[/tex]
Where Z is the Z-score,
X is the data point,
μ is the mean and σ is the standard variable.
As it is given that the mean of the normal population is 9, while the standard deviation is 5. Therefore,
A.) The proportion of the population is less than 20 can be written as,
[tex]\begin{aligned}P(X < 20)&=P(z < \dfrac{x-\mu}{\sigma})\\&=P(z < \dfrac{20-9}{5})\\&=0.9861\end{aligned}[/tex]
Hence, the proportion of the population is less than 20 is 0.9861
B.) The probability that a randomly chosen value will be greater than 6,
[tex]\begin{aligned}P(X > 6)&=1-P(z < \dfrac{x-\mu}{\sigma})\\\\&=1-P(z < \dfrac{6-9}{5})\\\\&=1-0.2743\\\\&=0.7257\end{aligned}[/tex]
Hence, the probability that a randomly chosen value will be greater than 6 is 0.7257.
Learn more about Z-score:
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