Respuesta :
The typical male wild mountain lion can weigh anywhere from 115 to 220 lbs (around 53 to 100 kg)
The average weight for the male lions is 62 kg
Meanwhile, typical female wild mountain lion can weigh around 64 - 141 lbs (around 29-64 kg) with the average of 42 kg.
The average weight for the male lions is 62 kg
Meanwhile, typical female wild mountain lion can weigh around 64 - 141 lbs (around 29-64 kg) with the average of 42 kg.
The sample mean is 91.8 lb and sample standard deviation is 31.4 lb.
Further explanation
How much do wild mountain lions weigh? Adult wild mountain lions (18 months or older) captured and released for the first time in the san andres mountains gave the following weights (pounds):
69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean weight x and sample standard deviation s. (Round your answers to one decimal place.)
x = lb
s = lb
(b) Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
lower limit lb
upper limit lb
.
a. Computation of sample mean and sample standard deviation
The number of samples, n = 6.
[tex]x = \frac{\sum X_i}{n} = \frac{69 +104+125+129+60+64}{6} \\x = 91.8\\s = \sqrt{\frac{\sum (x_i - x)^2}{n-1} } \\ s= \frac{(69-91.8)^2 + (104-91.8)^2 + (125-91.8)^2+(129-91.8)^2 + (60-91.8)^2 + (64-91.8)^2}{(6-1)} \\ s = 31.4[/tex]
Therefore the sample mean is 91.8 lb and sample standard deviation is 31.4 lb.
b. Confidence interval, when population standard deviation is not known
Confidence interval [tex]= x +- t_{\alpha/2} (\frac{s}{\sqrt{n} } )[/tex]
[tex]=( x - t_{(a/2),(n-1)} \frac{s}{\sqrt{n} } )< \mu < ( x + t_{(a/2),(n-1)} \frac{s}{\sqrt{n} } )[/tex]
Where,
- [tex]x = sample mean[/tex]
- [tex]t_{(a/2),(n-1)} \frac{s}{\sqrt{n} } = margin of error[/tex]
- [tex]( x - t_{(a/2),(n-1)} \frac{s}{\sqrt{n} } ) = lower limit of the mean[/tex]
- [tex]( x + t_{(a/2),(n-1)} \frac{s}{\sqrt{n} } ) = upper limit of the mean[/tex]
Computation of the 75% confidence interval for population mean is:
Where the sample mean (x-bar) is 91.8. The sample standard deviation (s) is 31.4. The confidence level is 0.75. Thus, the level of significance, α is [tex]0.25 (= 1 - 0.75)[/tex].
The degrees of freedom [tex](n - 1)[/tex] is [tex]5 (=6 - 1)[/tex]. For 75% confidence interval, the c
[tex](91.8 - (1.30 * \frac{31.4}{\sqrt{6} } )) < \mu < (91.8 + (1.30 * \frac{31.4}{\sqrt{6} } )) \\(91.8 - (16.66)) < \mu < (91.8 + 16.66)\\ 75.1 < \mu < 108.5[/tex]
Learn more
- Learn more about wild mountain lions https://brainly.com/question/12670494
- Learn more about the san andres mountains https://brainly.com/question/3150436
- Learn more about lions https://brainly.com/question/898658
Answer details
Grade: 9
Subject: biology
Chapter: lions
Keywords: wild mountain lions, the san andres mountains, lions, mountains san andres
