Garbage trucks entering a waste-management facility are weighed prior to offload

Question

Garbage trucks entering a waste-management facility are weighed prior to offloading their contents. The total processing time for a randomly selected truck at this facility is normally distributed with a mean of 13 min and a standard deviation of 4 min. a. What is the probability that a single truck's processing time is between 12 and 15 min? b. What is the probability that the average processing time for 16 trucks is between 12 and 15 min? Why is the probability in b. so much higher than the probability in a.? What is the probability that the average processing time for 16 trucks will be at least 20 min? c. d. It is desired to develop a confidence interval for a particular population with an assumed standard deviation of 3.0. Different confidence levels and different sample sizes are considered. a. b. c. d. Compute a 95% confidence interval the population mean when n 25 and x-58.3 Compute a 95% confidence interval the population mean when n 100 and x 58.3. Compute a 99% confidence interval the population mean when n-100 and x-583. How large must the sample size (n) be if the width of the 99% confidence interval is to be 1.0? Comment on the effect of increasing the sample size on the width of the confidence interval. Comment on the effect of increasing the confidence level of the interval. Do not simply state the impact-give an explanation for that impact.

Explanation / Answer

answering 4 parts as per cheg policies

mean 13

sd 4

a)

P(12<X<15)

= P(X<15)-P(X<12)

i know that, z= (X-mean)/sd

thus, P(12<X<15) = P(Z<(15-13)/4)-P(Z<(12-13)/4)

=P(Z<0.5)-P(Z<-0.25)

=0.290168787

b)

P(12<X<15)

=P(X<15)-P(X<12)

i know that, Z = (X-mean)/(sd/sqrt(n))

thus, P(12<X<15) = P(Z<(15-13)/(4/SQRT(16)) - P(Z<(12-13)/4/SQRT(16))

=P(Z<2) - P(Z<-1)

=0.818594614

c)

as the sample size increases, the z score increases. This increases the required probabiliy

d)

P(X>=20)

=1-p(x<=20)

i know that Z= (X-mean)/(sd/sqrt(n))

thus, P(X>=20) = 1 - P(Z<(20-13)/4/SQRT(16))

=1-p(z<7)

=1.27987E-12

=0.0000

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