Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random samples of 41 blended fuels are tested in a lab to ascertain the bio/total carbon ratio. (a) If the true mean is .9220 with a standard deviation of 0.0060, within what interval will 90 percent of the sample means fall? (Round your answers to 4 decimal places.)  The interval is from _______  to _________   (b) If the true mean is .9220 with a standard deviation of 0.0060, what is the sampling distribution of X⎯⎯⎯X¯ ?Exactly normal with μ = .9220 and σ = 0.0060.Approximately normal with μ = .9220 and σ = 0.0060.Exactly normal with μ = .9220 and σx⎯⎯=0.0060/41‾‾‾√.σx¯⁢= 0.0060/41.Approximately normal with μ = .9220 and σx⎯⎯=0.0060/41‾‾‾√.σx¯⁢= 0.0060/41.  1234 (c) What theorem did you use to answer part (b)?  Central Limit TheoremChebyshev's TheoremPythagorean TheoremLaw of Large Numbers

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Asked Sep 29, 2019
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Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random samples of 41 blended fuels are tested in a lab to ascertain the bio/total carbon ratio.
 
(a) If the true mean is .9220 with a standard deviation of 0.0060, within what interval will 90 percent of the sample means fall? (Round your answers to 4 decimal places.)
  
The interval is from _______  to _________  
 
(b) If the true mean is .9220 with a standard deviation of 0.0060, what is the sampling distribution of X⎯⎯⎯X¯ ?

  1. Exactly normal with μ = .9220 and σ = 0.0060.
  2. Approximately normal with μ = .9220 and σ = 0.0060.
  3. Exactly normal with μ = .9220 and σx⎯⎯=0.0060/41‾‾‾√.σx¯⁢= 0.0060/41.
  4. Approximately normal with μ = .9220 and σx⎯⎯=0.0060/41‾‾‾√.σx¯⁢= 0.0060/41.

  

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(c) What theorem did you use to answer part (b)?
  

  • Central Limit Theorem
  • Chebyshev's Theorem
  • Pythagorean Theorem
  • Law of Large Numbers

 

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Expert Answer

Step 1

It is given that the mean and standard deviation are 0.9220 and 0.0060, respectively. The sample size n is 41.

Step 2

For 90%, the z-score value using normal table is 1.645.

The...

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CI uzx Vn 0.9220±1.645y 0.0060 41 =0.9220t0.00 15 =(0.9220-0.0015,0.9220 +0.0015 =(0.9205,0.9235)

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