Final answer:
The answered questions relate to the central limit theorem, where the distribution of sample means and sums approach a normal distribution as the sample size increases. The questions address the calculation of mean and standard deviation for the sampling distributions derived from different population parameters.
Step-by-step explanation:
The questions pertain to the concept of the central limit theorem and the sampling distribution of both sample means and sums.
For the normal distribution X~ N(60, 9)
- The distribution of sample means ℓ (X-bar) for samples of size 25 is normally distributed with mean μ = 60 and standard deviation σ/√n = 9/5.
- The distribution of sample sums EX is also normally distributed with mean nμ = 25×60 and standard deviation nσ = 25×9.
For an unknown distribution with a mean of 12 and a standard deviation of one
- The mean of EX is nμ = 25×12.
For a random variable with a mean of 25 and a standard deviation of two
- The distribution for the sample mean of samples of size 100 is normally distributed with mean μ = 25 and standard deviation σ/√n = 2/10.
These examples demonstrate the application of the central limit theorem, which indicates as the sample size n increases, the distribution of the sample means and sums tend towards a normal distribution.