Question

Garbage trucks entering a particular waste-management facility are weighed prior to offloading their contents. Let X = the total processing time for a randomly selected truck at this facility (waiting, weighing, and offloading). An article suggests the plausibility of a normal distribution with mean 14 min and standard deviation 4 min for X. Assume that this is in fact the correct distribution. (Round your answers to four decimal places.) (a) What is the probability that a single truck's processing time is between 11 and 17 min? (b) Consider a random sample of 16 trucks. What is the probability that the sample mean processing time is between 11 and 17 min? (c) Why is the probability in (b) much larger than the probability in (a)? The mean is u = 14. A sample mean X based on n = 16 observations is likely to be ---Select-- the true mean than is a single observation X. That is because, while both are "centered" at u, the ---Select--- variability in X gives it --Select-- ability to vary significantly from u. (d) What is the probability that the sample mean processing time for a random sample of 16 trucks will be at least 23 min?

Answer 1

A.

P(11<X<17) = P((11-14)/4 < Z< (17-14)/4)

= P(-0.75<Z<0.75)

= 0.5467

B.

$\mu _{\bar{x}}$ = $\mu$ = 14

$\sigma _{\bar{x}}$ =
0/vn
= 1

P(11<<17) = P((11-14)/1 < Z< (17-14)/1)

X

= P(-3<Z<3)

= 0.9973

C.

far from

higher

more

D.

P(
X
$\geq$23) = P(Z$\geq$(23-14)/1)

= P(Z$\geq$9)

~ 0

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