Functionally important traits in animals tend to vary little from one individual to the next within populations, possibly because individuals who deviate too much from the mean have lower fitness. If this is the case, does variance in a trait rise after it becomes less functionally important? Billet et al. (2012) investigated this question with the semicircular canals (SC) of the three-toed sloth (Bradypus variegatus). The authors proposed that since sloths don't move their heads much, the functional importance of SC is reduced, and may vary more than it does in more active animals. They obtained the following measurements of the ratio of the length to width of the anterior SC in 6 sloths. Assume this represents a random sample. In other, more active animals, the standard deviation of this ratio is 0.09.

The 95% confidence interval for the standard deviation is (0.14, 0.54).

Step-by-step explanation:

The complete data and question is:

Sloth CW Ratios ; {1.5 , 1.09 , 0.98 , 1.42 , 1.49 , 1.25}

The 95% confidence interval for the standard deviation of this data is  < σ <   (two decimals - include the leading zero) .

Solution:

Compute the sample standard deviation as follows:

The (1 - α)% confidence interval for the variance is:

Confidence level = 95%

⇒ α = 0.05

The degrees of freedom is,

df = n - 1 = 6 - 1 = 5

Compute the critical values of Chi-square:

*Use a Chi-square table.

Compute the 95% confidence interval for the variance as follows:

Thus, the 95% confidence interval for the standard deviation is (0.14, 0.54).

the answer is here: : )

:

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x