Let #X# be #N(mu, sigma^2)# so that #P(X<89) = 0.90# and #P(X<94) = 0.95#. How do you find #mu# and #sigma^2#?

We find the associated value of #z# from the standard normal distribution #Z" ~ "N(0, 1)#, such that #"P"(X < x) = "P"(Z < z)#, using the relation #z = (x - mu)/sigma#, as follows:

#"P"(X < x) = "P"(Z < (x - mu)/sigma)#

so

#"P"(X<89) = "P"(Z < (89 - mu)/sigma)#
#"            "0.90 = "P"(Z < (89 - mu)/sigma)#

And from #z#-table lookup, the #z# value that gives #"P"(Z < z)= 0.90# is #z = 1.28#. Thus, we know

#z = 1.28 = (89 - mu) / sigma#

Similarly, we also use #"P"(X< 94) = 0.95# to get

#z = 1.65=(94 - mu)/sigma#

We now have a system of two equations in two unknowns: #mu# and #sigma#. We can solve this system as follows:

#{(1.28 sigma = 89 - mu),(1.65 sigma = 94 - mu):}#

#=> {(mu = 89 - 1.28 sigma),(mu = 94 - 1.65 sigma):}#

#=>89 - 1.28 sigma = 94 - 1.65 sigma#

#=>"         "0.37 sigma = 5#

#=>"                "sigma = 13.51#

so

#mu = 89 - 1.28 sigma#

#=>mu = 89 - 1.28(13.51)#
#color(white)(=> mu) = 89 - 17.30#
#color(white)(=> mu) = 71.70#

Finally, #sigma = 13.51# means #sigma ^2 = 13.51^2 = 182.52#.

Thus, we have #mu = 71.70# and #sigma = 182.52#.