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3 years ago

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a company pays its employees and average hourly wage of 36.25 with a standard deviation of 6.50 more than 23.22 of 36 employees

are randomly selected, what is the probability that their mean wage is more than 23.22

1 answer:

Yuliya22 [10]3 years ago

8 0

Answer:

P [ X > 23,22 ] = 0,8798 or P [ X > 23,22 ] = 87,98 %

Step-by-step explanation:

P [ X > 23,22 ] = 1 - P [X ≤ 23,22 ]

P [X ≤ 23,22 ] = ( X - μ₀ ) / σ /√n

P [X ≤ 23,22 ] = ( 23,22 - 36,25 )/ 6,50/√36

P [X ≤ 23,22 ] = - 13,03 *6 / 6,50

P [X ≤ 23,22 ] = - 12,02 or P [X ≤ 23,22 ] = 12,02 %

(note the - sign only mean that values on the left tail of the bell shape curve)

Then

P [ X > 23,22 ] = 1 - 0,1202

P [ X > 23,22 ] = 0,8798 or P [ X > 23,22 ] = 87,98 %

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Step-by-step explanation:

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Since sine is a bounded function between -1 and 1, the only solution that is mathematically reasonable is:

$\sin A \approx 0.382$

By means of inverse trigonometrical function, we get the value associate of the function in sexagesimal degrees:

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$f(A) = \sin 2A + \frac{1}{\sin 2A}$

$f(A) = \frac{\sin^{2} 2A+1}{\sin 2A}$

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<u>Direct variation equation:</u>

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