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Answer:
46.72%.
Step-by-step explanation:
We have been given that the number of boxes of supplies received weekly is normally distributed with a mean of 200 and a standard deviation of 20. We are asked to find the number of boxes of supplies that arrive will be greater than 210 or less than 180.
We will use z-score formula and normal distribution table to solve our problem.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{180-200}{20}[/tex]
[tex]z=\frac{-20}{20}[/tex]
[tex]z=-1[/tex]
Let us find z-score corresponding to normal score 210.
[tex]z=\frac{210-200}{20}[/tex]
[tex]z=\frac{10}{20}[/tex]
[tex]z=0.5[/tex]
[tex]P(z<-1)+P(z>0.5)=P(z<-1)+1-P(z<0.5)[/tex]
Using normal distribution table, we will get:
[tex]P(z<-1)+P(z>0.5)=0.15866+1-0.69146[/tex]
[tex]P(z<-1)+P(z>0.5)=0.4672[/tex]
[tex]0.4672\times 100\%=46.72\%[/tex]
Therefore, 46.72% of the time the number of boxes of supplies that arrive will be greater than 210 or less than 180.
The percentage of the time that the number of boxes of supplies that arrive will be greater than 210 or less than 180 is; 46.719%
We are given;
Population mean; μ = 200
Population standard deviation; σ = 20
We want to find the percentage of the time will the number of boxes of supplies that arrive be greater than 210 or less than 180. That is at;
Sample mean of x' = 180 and x' = 210.
Formula for z-score is;
z = (x' - μ)/σ
At x' = 180
z = (180 - 200)/20
z = -20/20
z = -1
At x' = 210
z = (210 - 200)/20
z = 10/20
z = 0.5
From online p-value calculator between two z-values, we have;
P(x < -1 or x > 0.5) = 0.46719
Or P(x < -1 or x > 0.5) = 46.719%
Read more about z-score at; https://brainly.com/question/25638875
