Answer :
(a) The sample mean is 98.23.
(b) The standard deviation is 16.97.
(c) The maximum likelihood estimate is 264.24.
(d) An unbiased estimate is 288.26.
(e) The sample median is 96.3
Bill Montana went into the mountains of Idaho and captured 12 random mountain lions. Their weights, in pounds, were the following:
93.0, 109.4, 88.2, 114.5, 83.0, 103.2, 110.7, 128.9, 62.8, 97.9, 94.7, 92.5
a) We have to find the sample mean.
Sample mean = Sum of all weights/ Total number
Sample mean = 1178.8/12
Sample mean = 98.23
b) We have to find the standard deviation.
The formula is
Standard Deviation(SD) = √{∑(Mean - Weight)^2/ Total number}
After using the formula
Sample standard deviation (SD) = 16.97
c) We have to calculate the maximum likelihood estimate.
The MLE of sigma^2 is given by:
sigma^2_(MLE) = 1/nsum_(i = 1)(x_i-barx)^2
by calculations we get ,
sigma^2_(MLE) = 264.24
d) We have to calculate an unbiased estimate.
The unbiased estimator of sigma^2 " is"
sigma^2_(UE) = 1/(n-1)sum_(i = 1)(x_i-barx)^2
By calculations, we get:
sigma^2_(UE) = 288.26
e) Now we have to calculate the sample median.
The sample median = 96.3.
h) Now we have to calculate the 90th percentile of a standard normal distribution.
let, the 90th percentile of standard normal distribution be "k"
then ,
P(X < k) = 0.90
=> P(X < k) = P(X < 1.281)
=> k = 1.281
Therefore, the 90th percentile of standard normal distribution is 1.281.
(i) Now we have to find an 80% confidence interval.
The upper bound of confidence interval of is by calculations , we get 101.91.
The lower bound of confidence interval of is by calculations, we get 94.54.
Therefore, the confidence interval is : (94.54 , 101.91)
To learn more about confidence interval link is here
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