L. Michael V. answered 11/25/20
Former Actuarial Analyst
The distribution of X is a normally distributed variable with mean 65 million and standard deviation of 18 million. This is the population distribution.
The distribution of x bar is a normally distributed variable with a mean of 65 million and a standard deviation of 18 million/sqrt(7). This is the sampling distribution.
To find parts c and d, you can use the normalcdf command in your TI84 calculator (or convert the lower and upper bounds to z-scores and use a table)
C: normalcdf(lowerbound = 61, upperbound = 64, mean = 65, sd = 18) = 0.0658
D: C: normalcdf(lowerbound = 61, upperbound = 64, mean = 65, sd = 18/sqrt(7)) = 0.1633
The assumption of normality is not necessary for part D because it was given that the population was normally distributed, thus the sampling distribution of sample size 7 is automatically normally distributed as it flows from the fact that the population is normally distributed.
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