An animal trainer obtained the following sample data (Table A) in a study of reaction times of dogs (in seconds) to a specific stimulus. He then selected another group of dogs that were much older than the first group and measure their reaction times to the same stimulus. The sample data is shown in Table B.

Question

An animal trainer obtained the following sample data (Table A) in a study of reaction times of dogs (in seconds) to a specific stimulus. He then selected another group of dogs that were much older than the first group and measure their reaction times to the same stimulus. The sample data is shown in Table B.

Table A Table B
Classes Frequency Classes Frequency
2.3-2.9 24 2.3-2.9 4
3.0-3.6 24 3.0-3.6 21
3.7-4.3 14 3.7-4.3 8
4.4-5.0 5 4.4-5.0 0
5.1-5.7 4 5.1-5.7 24
5.8-6.4 16 5.8-6.4 20

Find the variance and standard deviation for the two distributions above. Compare the variation of the data sets. Decide if one data set is more variable than the other

Answer 1

Copy and paste does not work. Even if we had the tables, we don't know what your question is. Repost with more adequate data.

Answer 2

To find the variance and standard deviation for the two distributions, we need to calculate the mean of each distribution first.

To find the mean, we multiply each value in the Classes column by its corresponding frequency, then sum up all these values, and divide by the total frequency.

For Table A:
Mean_A = (24 * 2.6 + 24 * 3.3 + 14 * 4.0 + 5 * 4.7 + 4 * 5.4 + 16 * 6.1) / (24 + 24 + 14 + 5 + 4 + 16) = 3.734

For Table B:
Mean_B = (4 * 2.6 + 21 * 3.3 + 8 * 4.0 + 0 * 4.7 + 24 * 5.4 + 20 * 6.1) / (4 + 21 + 8 + 0 + 24 + 20) = 4.946

Next, we calculate the variance for each distribution.

For Table A:
Variance_A = [(24 * ((2.6 - 3.734)^2) + 24 * ((3.3 - 3.734)^2) + 14 * ((4.0 - 3.734)^2) + 5 * ((4.7 - 3.734)^2) + 4 * ((5.4 - 3.734)^2) + 16 * ((6.1 - 3.734)^2)) / (24 + 24 + 14 + 5 + 4 + 16)] = 1.300

For Table B:
Variance_B = [(4 * ((2.6 - 4.946)^2) + 21 * ((3.3 - 4.946)^2) + 8 * ((4.0 - 4.946)^2) + 0 * ((4.7 - 4.946)^2) + 24 * ((5.4 - 4.946)^2) + 20 * ((6.1 - 4.946)^2)) / (4 + 21 + 8 + 0 + 24 + 20)] = 1.715

Finally, we calculate the standard deviation for each distribution by taking the square root of the variance.

Standard Deviation_A = √1.300 ≈ 1.14
Standard Deviation_B = √1.715 ≈ 1.31

To compare the variation of the data sets, we can look at the standard deviations. A higher standard deviation indicates a greater variation in the data set.

In this case, the standard deviation for Table B (1.31) is larger than the standard deviation for Table A (1.14), which suggests that the data set for Table B (older dogs) has a greater variation than the data set for Table A (younger dogs).