Answer :
The sample mean weight is x = 91.5000 lb, and the sample standard deviation is s ≈ 30.0639 lb and the 75% confidence interval for the population average weight of all adult mountain lions in the specified region is approximately 71.8 lb to 111.2 lb.
(a) The weights of adult wild mountain lions have an approximately normal distribution. Therefore, the sample mean and sample standard deviation can be calculated as follows:
First, calculate the sum of the given weights:65 + 106 + 126 + 128 + 60 + 64 = 549
Then, calculate the sample mean: x = 549/6 ≈ 91.5 lb Round x to four decimal places: x = 91.5000 lb
Finally, calculate the sample standard deviation using the formula
:s = sqrt [ Σ ( xi - x )2 / ( n - 1 ) ]
where xi represents each weight, and n is the sample size:
s = sqrt [ ( (65 - 91.5)2 + (106 - 91.5)2 + (126 - 91.5)2 + (128 - 91.5)2 + (60 - 91.5)2 + (64 - 91.5)2 ) / ( 6 - 1 ) ]≈ 30.0639 lb Round s to four decimal places: s ≈ 30.0639 lb
Therefore, the sample mean weight is x = 91.5000 lb, and the sample standard deviation is s ≈ 30.0639 lb.
(b) To find the 75% confidence interval for the population average weight of all adult mountain lions, we need to use the formula:
x ± z(α/2) * (s / sqrt(n))
where: x is the sample mean weight z(α/2) is the z-score that corresponds to the desired confidence level.
Since we want a 75% confidence interval, α = 1 - 0.75 = 0.25, and z(α/2) ≈ 0.6745s is the sample standard deviation is the sample size Plugging in the given values, we get:
x ± z(α/2) * (s / sqrt(n))≈ 91.5000 ± 0.6745 * (30.0639 / sqrt(6))≈ 91.5000 ± 19.7282 lb
Round the lower and upper limits to one decimal place: lower limit ≈ 71.8 lb upper limit ≈ 111.2 lb
Therefore, the 75% confidence interval for the population average weight of all adult mountain lions in the specified region is approximately 71.8 lb to 111.2 lb.
to learn more about confidence intervals refer to:
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