a. Let
be the CDF of
. The CDF of
is

which is to say,
is also normally distributed, but with different parameters. In particular,
![E[Y]=E[5X+2]=5E[X]+2=17](https://img.qammunity.org/2020/formulas/mathematics/college/kh0y9sbio55za08jntymt4sx5v1vwck5vk.png)
![\mathrm{Var}[Y]=\mathrm{Var}[5X+2]=5^2\mathrm{Var}[X]=100](https://img.qammunity.org/2020/formulas/mathematics/college/qkxf46c7219i1b1cr0saslvo5a6y266tc2.png)
b. Using the appropriate CDFs, we have


c. The 99th percentile for any distribution
is the value of
such that
, i.e. all values of
below
make up the lower 99% of the distribution.
We have

d. On the other hand, the 99th percentile for
is

e. We have

which suggests that
is normally distributed, or
is log-normally distributed. Recall that the moment-generating function for
is

But we also have
![M_Y(t)=E[e^(tY)]=E[e^(t\ln W)]=E[W^t]](https://img.qammunity.org/2020/formulas/mathematics/college/34wlnjhhiauebmt3rwdb3s5m86qia9bh1o.png)
Then
![E[W]=M_Y(1)=e^(67)](https://img.qammunity.org/2020/formulas/mathematics/college/hz3e9roetlv4n98in8qncfvy3a2i27i6mu.png)
and
![E[W^2]=M_Y(2)=e^(234)\implies\mathrm{Var}[W]=E[W^2]-E[W]^2=e^(234)-e^(134)](https://img.qammunity.org/2020/formulas/mathematics/college/t5q0ezrgn6rnk6dv6x7qxzasxc36q6xx6u.png)