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Compare 5/6 and 1/3 using the benchmark fraction 1/2
1 answer:
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Part A:
From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.
Part B:
The standard deviation is given by the population deviation divided by the square root of the sample size.
Part C:
The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by
Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that <span>the average number of moths in 30 traps is greater than 0.7</span> given by:
From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.
Part B:
The standard deviation is given by the population deviation divided by the square root of the sample size.
Part C:
The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by
Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that <span>the average number of moths in 30 traps is greater than 0.7</span> given by: