in this exercise we will explore the link between the standard normal distribution, Z ~ N(mean=0, variance=1), and the Student's t distribution, X ~ Student's t(d.o.f.=n-1). Calculate the following using the Excel function =NORM.INV or =T.INV as appropriate.

1. Calculate the values of Z, where Z ~ N(mean=0, variance=1), for which 0.5% of the data lies in the lower tail and 0.5% lies in the upper tail.

(answer in absolute value and to 2 decimal places)

2.Calculate the values of X, where X ~ Students t(d.o.f.=n-1), for which 0.5% of the data lies in the lower tail and 0.5% lies in the upper tail, and where n=10 such that X ~ Students t(d.o.f.=9).

(answer in absolute value and to 2 decimal places)

3.Calculate the values of X, where X ~ Students t(d.o.f.=n-1), for which 0.5% of the data lies in the lower tail and 0.5% lies in the upper tail, and where n=20 such that X ~ Students t(d.o.f.=19).

(answer in absolute value and to 2 decimal places)

4.Calculate the values of X, where X ~ Students t(d.o.f.=n-1), for which 0.5% of the data lies in the lower tail and 0.5% lies in the upper tail, and where n=50 such that X ~ Students t(d.o.f.=49).

(answer in absolute value and to 2 decimal places)

5.Calculate the values of X, where X ~ Students t(d.o.f.=n-1), for which 0.5% of the data lies in the lower tail and 0.5% lies in the upper tail, and where n=100 such that X ~ Students t(d.o.f.=99).

(answer in absolute value and to 2 decimal places)

could you explain these things?