Please help me with this question with explanation (I really appreciate if you can help me the procedure to use T83 to calculate this). Thank you so much!

Question 1 1 pts A sampling distribution is a distribution of size n where each value is the mean of a sample of size n. The central limit theorem tells us some important information about the sampling distributions: . For large enough n, the sampling distributions are approximately normal. Large enough is often as small as 30, but if a population is very skewed it is possible a larger sample size is needed. . The mean of the sampling distribution (the mean of all the means for all the possible samples of size n from the population) is equal to the mean of the population. Hi = / . The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size. of = Vn A sample of 62 people was taken from a population that has a mean of 134 and a standard deviation of 17. What is the mean of the sampling distribution of size 62?Question 2 1 pts A sampling distribution is a distribution of size n where each value is the mean of a sample of size n. The central limit theorem tells us some important information about the sampling distributions: . For large enough n, the sampling distributions are approximately normal. Large enough is often as small as 30, but if a population is very skewed it is possible a larger sample size is needed. . The mean of the sampling distribution (the mean of all the means for all the possible samples of size n from the population) is equal to the mean of the population. HI = / . The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size. of = in A sample of 149 people was taken from a population that has a mean of 25.8 and a standard deviation of 44. What is the standard deviation of the sampling distribution of size 149?Question 3 1 pts When working on a normal distribution other than the standard normal distribution you need to either first convert to the z-scores and then find probabilities using the standard normal distribution(N(0,1)) OR when using technology you can input the mean and standard deviation. P (a