Building a railroad to connect 2 train stations 100 kilometres apart. We are con

Question

Building a railroad to connect 2 train stations 100 kilometres apart. We are concerned that the factory making the rail sections is making them shorter than the desired 25m length. We took a random sample of 10 rail sections, measured their lengths (in metres) and obtained the following data:

*** Need to know how to do the procedures for B, C, and D

C4 25.0150 24.9830 24.9899 25.0081 24.9809 T6 C7 Tg C10 24.9919 24.9891 249801 24.9859|24.9881 (a) Compute the point estimate of the population average rail length (Hint: (b) Our plans call for 4000 rail sections per side to connect the two stations. (c) The manufacturer assures us that the population mean is = 25what 22:i = 249.9120) If we use the rails as they are, what will be distance shortfall? is the probability that a sample of size 10 would have a mean length of at most the length found in part (a)? (We know that the population = 0.00839) Do you think the manufacturer is correct? Explain (d) We need the rails to be (25 ± 0.015)m. Given our current production output, estimate the proportion of rails that do not meet our requirement

Explanation / Answer

Part a

We are given

Xi = 249.9120, n = 10

Point estimate = Xi / n = 249.9120/10 = 24.9912

Part b

If we use the rails as they are, then shortfall distance = 25m per rail section

For 4000 rail section, total shortfall distance = 25*4000 = 100000 meters

Required answer = 100000 meters

Part c

We are given

µ = 25

Xbar = 24.9912

= 0.00839

n = 10

We have to find P(Xbar<24.9912)

Z = (Xbar - µ) / [/sqrt(n)]

Z = (24.9912 – 25) / [0.00839/sqrt(10)]

Z = -0.0088/0.002653151

Z = -3.316810841

P(Z<-3.316810841) = P(Xbar<24.9912) = 0.000455256

(By using z-table or excel)

Required probability = 0.000455256

Manufacturer is correct because probability is unusual (less than 0.05.)

Part d

Here, we have to find the probability outside the given interval.

That is we have to find

1 – [P(25 – 0.015 < X < 25 + 0.015)] = 1 – P(24.985<X<25.015)

P(24.985<X<25.015) = P(X<25.015) – P(X<24.985)

For X = 25.015

Z = (X – µ) /

Z = (25.015 - 25) / 0.00839

Z = 1.78784267

P(Z<1.78784267) = P(X<25.015) = 0.963099303

(By using z-table or excel)

For X = 24.985

Z = (X – µ) /

Z = (24.985- 25) / 0.00839

Z = -1.78784267

P(Z<-1.78784267) = P(X<24.985) = 0.036900697

(By using z-table or excel)

P(24.985<X<25.015) = P(X<25.015) – P(X<24.985)

P(24.985<X<25.015) = 0.963099303 - 0.036900697

P(24.985<X<25.015) = 0.926198606

1 – [P(25 – 0.015 < X < 25 + 0.015)] = 1 – P(24.985<X<25.015)

1 – [P(25 – 0.015 < X < 25 + 0.015)] = 1 – 0.926198606

1 – [P(25 – 0.015 < X < 25 + 0.015)] = 0.073801394

Required Probability = 0.073801394

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