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lesantik [10]
3 years ago
12

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 sele

cted 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?
Mathematics
1 answer:
Yuliya22 [10]3 years ago
6 0

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

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Step-by-step explanation:

The cylindrical part

Area of the cylindrical part = Circumference of the circle * height

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I know this doesn't look right according to the diagram, but it is the only way that you can solve the question. That is, you must assume there are two lids, not one as the diagram suggests.

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3 years ago
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