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Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article reports that fifteen tracks on one rock specimen had an average track length of 12 μm with a standard deviation of 2 μm. Assuming this to be a random sample from an approximately normal population, find a 99% confidence interval for the mean track length for this rock specimen. Round the answers to three decimal places.

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Answer:

Mean track length for this rock specimen is between 10.463 and 13.537

Step-by-step explanation:

99% confidence interval for the mean track length for rock specimen can be calculated using the formula:

M±[tex]\frac{t*s}{\sqrt{N}}[/tex] where

  • M is the average track length (12 μm) in the report
  • t is the two tailed t-score in 99% confidence interval (2.977)
  • s is the standard deviation of track lengths in the report (2 μm)
  • N is the total number of tracks (15)

putting these numbers in the formula, we get confidence interval in 99% confidence as:

12±[tex]\frac{2.977*2}{\sqrt{15}}[/tex] =12±1.537

Therefore, mean track length for this rock specimen is between 10.463 and 13.537

adefioyelaoye

The mean track length for this rock specimen is between 10.463 and 13.537.

What is Confidence Interval?

This refers to the probability that a population parameter will fall between two set values.

We were told that there was 99% confidence interval for the mean track length for rock specimen. It can then be calculated using the formula:

M± (t×s/√N)

where

M = average track length = 12 μm

t = two tailed t-score in 99% confidence interval = 2.977

s = standard deviation =2 μm

N = total number of tracks = 15

We can substitute the parameters into the formula

12± (2.977× 2/√15) = 12±1.537

The mean track length for this rock specimen is between 10.463 and 13.537.

Read more about Confidence Interval here https://brainly.com/question/15712887

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