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According to a 2016 survey, 6 percent of workers arrive to work between 6:45 A.M. and 7:00 A.M. Suppose 300 workers will be selected at random from all workers in 2016. Let the random variable W represent the number of workers in the sample who arrive to work between 6:45 A.M. and 7:00 A.M. Assuming the arrival times of workers are independent. What is closest to the standard deviation of W?

Respuesta :

Answer:

Closest standard deviation of W = 4.11

Step-by-step explanation:

Randomly selected workers = n =300

Probability of workers arriving between 6:45 to 7:00 A.M = p = 6% = 0.06

According to binomial distribution:

mean = μ = n.p

                = 300(0.06) = 18

variance = σ² = n.p(1-p)

                      = (300)(0.06)(1-0.06)

                      = (18)(0.94)

                      = 16.92

standard deviation = σ = 4.11

The Closest standard deviation of W = 4.1

Given that ;

Number of Randomly selected workers = n = 300

Probability of workers arriving between 6:45 to 7:00 A.M = p = 6% = 0.06

According to binomial distribution:

Mean = μ = n.p

                = 300(0.06)

               = 18

Then ;

variance = σ² = n.p(1-p)

                     = (300)(0.06)(1-0.06)

                     = (18)(0.94)

                    σ²  = 16.92

    Taking square root both sides.

                     σ = 4.1

As per question random variable W represent the number of workers who arrived at  6:45 A.M. and 7:00 A.M.

       W = 4.1

The closest standard deviation of W = 4.1

For more details about random binominal probability click the link given below.

https://brainly.com/question/16628500

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