The percent of stormtroopers that earn between 30,000 and 45,000 imperial credits annually is 38.117%.
The z score of a measure X in a collection with mean μ and standard deviation σ is given by:
Z = ( X - μ) / σ
We have,
The mean, μ = 36000
The standard deviation, σ = 15000
This is the p-value of Z when X = 30000 subtracted by the p-value of Z when X = 45000. So
X = 30000
Z = ( X - μ) / σ
Z = ( 30000 - 36000 ) / 15000
Z = - 6000 / 15000
Z = - 0.4
Using the z table, the p-value of 0.34458.
When X = 45000
Z = ( X - μ) / σ
Z = ( 45000 - 36000 ) / 15000
Z = 9000 / 15000
Z = 0.6
The p-value is 0.72575.
Therefore, the percentage of stormtroopers who earn between 30,000 and 45,000 imperial credits annually will be:
= 0.72575 - 0.34458
= 0.38117 = 38.117%
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