Question

Suppose a researcher used an electronic database to collect body weight (lb) for addolescent participants. The distribution of body weights (lbs) was categorized into three groups: Group-1, Group2 and Group- 3 based on the percentile. The researcher randomly selected participants by group: 14% from Group-1,76% from Group-2 and 10% from Group-3. The mean body weights of 100 sample participants were 101lb. Because the sample participants were selected by various groups from a big database system, researcher realized that sample variation of body weights were apparent for each group. The exact mean and standard deviation of the body weight by group are summary statistics. however the boundary of the mean body weight (lb) for each group is the most important summary. According to the database, the estimates of body weights are as following: standard deviation for Group-2 is equal to standard deviation for Group-3, and standard deviation for Group-2 is four times standard deviation for Group-1; coefficient of variation, CV (defined as standard deviation divided by mean ) for Group-1 was 0.05, and coefficient of variation for Group 2 = coefficient of variation for Group 3=0.09. Find out the 95% Chebyshev interval around the mean body weight for each group.

Answer 1

To find the 95% Chebyshev interval around the mean body weight for each group, use the formula: mean ± k * (CV * mean).

Step-by-step explanation:

To find the 95% Chebyshev interval around the mean body weight for each group, we need to use the coefficient of variation, which is defined as the standard deviation divided by the mean. The formula for the Chebyshev interval is: mean ± k * (CV * mean), where k is the number of standard deviations from the mean that includes at least (1 - 1/k²) of the data.

For Group-1, the coefficient of variation is 0.05, so the Chebyshev interval is: 101 ± k * (0.05 * 101).

For Group-2 and Group-3, the coefficient of variation is 0.09, so the Chebyshev interval is: 101 ± k * (0.09 * 101).