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Sagot :
The probability of winning exactly 21 times is:-0.32
Given,
The probability of winning on an arcade game, p = 0.659
Number of times you play arcade game, n = 30
By assuming this as a normal distribution, we get
Mean=μ=30×0.659 =19.77
Standard deviation, σ = [tex]\sqrt{np(1-p)}[/tex]
= [tex]\sqrt{30(0.659)(1-0.659)}[/tex]
≈ 2.60
Let X be a binomial variable
Then, the z score for x = 21 will be:
z = (x - μ) / σ = [tex]\frac{21-19.77}{2.60}[/tex]
≈ 0.47
Now, the probability of winning exactly 21 times
P (x ≥ z) = P (21 ≥ 0.47) = 0.32
Hence the probability of winning exactly 21 times is 0.32
Learn more about probability here:-https://brainly.com/question/13067945
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