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) ?

Question

Mathematics

Morton

3 April, 01:42

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) ?

+1

Answers (1)

Know the Answer?

Ali Duncan · Answer 1

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