Question

In a group of 10 black bears, the length of the longest claw was measured in each bear. These lengths are listed below in cm. What is the variance of these lengths? 2.8,3.1,3.8,3.2,3.3,2.9,3.2,3.1,3.5,3.0

(a).3.19
(b).3.08
(c).3.11
(d).3.30
(e).3.16

Answer 1

The variance of the lengths of the black bears' claws, calculated step by step, is found to be 0.0401 cm^2. The provided answer options do not match this calculated variance, suggesting a potential error in the provided options or the calculation.

Step-by-step explanation:

To determine the variance of the claw lengths of the black bears, we first need to calculate the mean (average) length. We then use this mean to find the sum of the squared differences between each bear's claw length and the mean. Finally, we divide this sum by the number of data points minus one to find the variance.

1. Calculate the mean of the lengths: (2.8 + 3.1 + 3.8 + 3.2 + 3.3 + 2.9 + 3.2 + 3.1 + 3.5 + 3.0) / 10 = 31.9 / 10 = 3.19 cm.
2. Calculate each length's deviation from the mean, square it, and sum these values:
   
   (2.8 - 3.19)2 + (3.1 - 3.19)2 + ... + (3.0 - 3.19)2 = 0.1521 + 0.0081 + 0.3721 + ... + 0.0361.
3. Sum of squared deviations: After calculating each, the sum is 0.361.
4. Divide the sum of squared deviations by n-1 (Degrees of Freedom): 0.361 / (10 - 1) = 0.0401 cm2.

The variance of the lengths of the black bears' claws is 0.0401 cm2.

None of the provided answers (a) 3.19, (b) 3.08, (c) 3.11, (d) 3.30, or (e) 3.16 match the calculated variance. It seems there might be a mistake in either the calculation or the provided options. Assuming correct calculations, all options are incorrect.