Question

a) Explain briefly what this means in practice that the length of the pipes is expected to be µ = 12. What is the probability that a randomly selected pipe is longer than 12.2 meters? What is the probability that the length of a randomly selected pipe is between 11.9 and 12.1 meters? What length is 90% of the tubes longer than? The company has been commissioned to produce a 10.8 kilometer (ie 10800 meters) long pipeline. The has a margin of error of ± 5 meters, that is, they can deliver a pipeline that is up to 5 meters shorter or 5 meters longer than the specified length of 10800 meters. The company is thinking of producing 900 pipes to the pipeline, and the total length of the pipeline then becomes the sum of the lengths of these pipes. Suppose each tube produced by the machine to the company has a length that is independent of the others tubes. b) What is the expected total length of the pipeline? What is the variance of the total length? What is the probability that the pipeline will be too long or too short (ie get a length beyond the allowed margin of error)? If the standard deviation of the lengths of the pipes, σ, could be adjusted to a different value, which value had to be adjusted so that the likelihood of the pipeline becoming too long or too long card to be 0.01? Before production is fully commissioned, one of the engineers at the company insists that they have to examine if the machine is properly adjusted so that the expected length of the pipes is really 12 meters. To examine this, they produce 12 tubes and measure their length. The results of the measurements are given below. We assume in the rest of the task that µ is unknown, while we still have that σ = 0.1 and that the lengths of different pipes are independent. M˚aleresultater: 11.87 11.82 11.99 12.01 11.93 11.98 12.08 12.11 11.92 11.79 12.02 12.07 c) Calculate the mean and median of the data. Find a 95% confidence interval for µ. How many measurements must be made in this situation at least

Answer 1

a) (i)

That means average length of pipes produced by the company is 12 meters

Let X be the length of a pipe with and

then

to find (ii)P( X >12.2)

(iii) P(11.9<X<12.1) = P(-1<Z<1) = 0.6816

(iv) P(Z>C) =0.90  

C = - 1.29

that means X = -1.29 *0.1 + 12 = 11.871

90% of the tubes longer than 11.87 meters

(b) (i)expected length of 900 pipelines = 900*12 = 10800 meters

(ii) variance = 900² * (0.1)² = 8100

then by central limit theorem ,

then

(iii) margin of error =

P( X< 10795 ) or P(X>10805) = P( Z< -0.06) + P(Z> 0.06) = 0.9522

Probability that length is too short or too long = 0.9522

(iv) P(Z<-C) + P(Z>C) = 0.01

C = 2.56 that is that is

standard deviation could be adjusted to 1.95

c) mean = (11.87 +...+12.07)/12 = 11.97 meters

median is the mean of 6th and 7th observation

median = 11.985 meters

95% CI for

=

=

=(11.91, 12.03)

Atleast 30 observation should have been made to use z value in CI

Note : for 95% CI , z = 1.96