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17 February, 09:17

Vending Machines Quarters are now manufactured so that they have a mean weight of 5.670 g and a standard deviation of 0.062 g, and their weights are normally distributed. A vend-ing machine is configured to accept only those quarters that weigh between 5.550 g and 5.790 g.

a. If 1 randomly selected quarter is inserted into the vending machine, what is the probability that it will be accepted?

b. If 4 randomly selected quarters are inserted into the vending machine, what is the probability that their mean weight is between 5.550g and 5.790g?

c. If you own the vending machine, which result is more important: the result from part (a) or part (b) ?

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Answers (1)
  1. 17 February, 11:14
    0
    Step-by-step explanation:

    mean = 5.67

    std. dev. = 0.062

    a)

    Probability that a randomly selected quarter weighs betwwen 5.55 and 5.79

    P (5.55 < X < 5.79)

    = P ((5.55 - 5.67) / 0.062) < X < ((5.79 - 5.67) / 0.062)

    = P (-1.9355 < z < 1.9355)

    = P (z < 1.9355) - P (z < - 1.9355)

    = 0.9735 - 0.0265

    = 0.9471

    Probability of such event is 0.9471

    b)

    SE = 0.062/sqrt (4) = 0.031

    P (5.55 < X < 5.79)

    = P ((5.55 - 5.67) / 0.031) < X < (5.79 - 5.67) / 0.031)

    = P (-3.8709 < z < 3.8709)

    = P (z < 3.8709) - P (z < - 3.8709)

    = 0.9999 - 0.000054

    = 0.99989

    Probability is 0.99989

    c)

    Results from part (a) are more important
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