An unknown distribution has a mean of 44 and a standard deviation of 7.4. A sample size of 68 is drawn randomly from the population. Find the sum that is 0.96 standard deviations above the mean of the sums. State your answer to two decimal places.

The mean of the sampling distribution of the sum of a random sample can be found by: μ_sum = n * μ where n is the sample size, μ is the population mean, and μ_sum is the mean of the sum of the sample. The standard deviation of the sampling distribution of the sum of a random sample can be found by: σ_sum = sqrt(n) * σ where n is the sample size, σ is the population standard deviation, and σ_sum is the standard deviation of the sum of the sample. Using the given information: μ = 44 σ = 7.4 n = 68 Therefore: μ_sum = n * μ = 68 * 44 = 2992 σ_sum = sqrt(n) * σ = sqrt(68) * 7.4 ≈ 52.22 To find the sum that is 0.96 standard deviations above the mean of the sums, we can use the following formula: x = μ_sum + z * σ_sum where x is the desired sum, z is the number of standard deviations from the mean, μ_sum is the mean of the sums, and σ_sum is the standard deviation of the sums. Substituting the given values: z = 0.96 x = 2992 + 0.96 * 52.22 ≈ 3086.38 Therefore, the sum that is 0.96 standard deviations above the mean of the sums is approximately 3086.38.