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Use the given values of nequals=93 and pequals=0.24 to find the maximum value that is significantly​ low, muμminus−2sigmaσ​, and the minimum value that is significantly​ high, muμplus+2sigmaσ. Round your answer to the nearest hundredth unless otherwise noted.

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Answer:

The maximum value that is significantly low is 14.0828

The minimum value that is significantly​ high is 30.5572

Step-by-step explanation:

If we assume a binomial distribution, n is equal to 93 and p is equal to 0.24, the mean μ and the standard deviation σ are calculated as:

μ = n*p = 93*0.24 = 22.32

σ = [tex]\sqrt{n*p*(1-p)} =\sqrt{93*0.24*(1-0.24)} =4.1186[/tex]

Then, the maximum value that is significantly​ low, μ−2σ, and the minimum value that is significantly​ high, μ+2σ, are equal to:

μ − 2σ = 22.32 - 2(4.1186) = 14.0828

μ + 2σ = 22.32 + 2(4.1186) = 30.5572

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The minimum value that is significantly​ high is 30.56 and the maximum value that is significantly low is 14.08.

Minimum and maximum value

First step is to find the mean using this formula

Mean (μ)=n×p

Where:

n=93

p=0.24

Mean (μ)= 93×0.24

Mean (μ)= 22.32

Second step is to find the standard deviation

Standard deviation(σ)=√n×p ×(1-p)

Standard deviation(σ)=√93×0.24×(1-0.24)

Standard deviation(σ)=4.1186

Third step is to find the maximum and minimum value

Minimum value=μ − 2σ

Minimum value= 22.32 - 2(4.1186)

Minimum value=14.08

Maximum value=μ + 2σ

Maximum value= 22.32 + 2(4.1186)

Maximum value= 30.5572

Maximum value=30.56(Approximately)

Inconclusion the minimum value that is significantly​ high is 30.56 and the maximum value that is significantly low is 14.08.

Learn more about minimum and maximum value here:https://brainly.com/question/14993153

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