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aalyn [17]
3 years ago
15

In a survey of employee satisfaction, 60% of the employees are male and 45% of the employees are satisfied. What is the probabil

ity of randomly selecting an employee who is male and satisfied?Immersive Reader (10 Points)
SAT
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

The probability of being a male and a satisfied employee is 27%

Explanation:

The probability of being a male P(m) = 60% = 60/100 = 0.6

The probability of being a satisfied employee P(s) = 45% = 45/100 = 0.45

Mathematically, the probability of being a male and satisfied = Probability of being a male * Probability of being a satisfied employee = P(m) * P(s) = 0.6 * 0.45 = 0.27

or simply 27/100 which is same as 27%

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Based the density function of the observed outcome, the probability P(x\leq \frac{1}{3} ) is equal to 0.1388.

<h3>What is a density function?</h3>

A density function can be defined as a type of function which is used to represent the density of a continuous random variable that lies within a specific range.

Given the following density function:

f(x)=\left \{ {{2(1-x),\;0 < x < 1} \atop {0,\; otherwise}} \right.

Next, we would calculate P(x\leq \frac{1}{3} );

P(x\leq \frac{1}{3} )=\int\limits^\frac{1}{3} _0 {f(x)} \, dx \\\\P(x\leq \frac{1}{3} )=2 \int\limits^\frac{1}{3} _0 {(1-x)} \, dx \\\\P(x\leq \frac{1}{3} )=2|x-\frac{x^2}{2} |\limits^\frac{1}{3} _0\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{2} (\frac{1}{3})^2]\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{18}]\\\\P(x\leq \frac{1}{3} )=0.1388

Read more on probability here: brainly.com/question/25870256

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2 years ago
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