normal distribution finding b For part b, I wrote the answer like this:
Pr(− < ( − )/ < ) = 0.90
Pr(− < Z < ) = 0.90
b=（1.64＋1.65)/2 =1.645
Am I correct doing like this? I'm not sure.
Do I need to do the inversion of centralise by making it times  and plus ?
Question: If random variable  follows Normal distribution
(a)If we know () = 75 and () = 100 please find Pr(X < 60) and
Pr(70 < X < 100)
(b)  Find  so that Pr(− < ( − )/ < ) = 0.90
 A: b)
the request is the same as
$$\mathbb{P}[Z<b]=0.95$$
Reading the result on Z-table you get
$$b\approx 1.64$$
this because you have to calculate the two quantiles excluding 5% on the left tail and another 5% on the right one.
A: If you use R (or other software) you typically don't
need to standardize in order to solve such problems.
Here are answers in R: The second parameter is the standard deviation $\sigma = 10.$
If $X \sim \mathsf{Norm}(\mu=75, \sigma=10),$ then $P(70 < X < 100).$
diff(pnorm(c(70,100), 75, 10))
[1] 0.6852528
pnorm(100, 75, 10) - pnorm(70, 75, 10)
[1] 0.6852528

Find $b$ such that $P(− < Z < ) = 0.90.$ By symmetry, this amounts to
finding $P(Z < b) = 0.95.$ Then $b = 1.644854$ often rounded to $1.645.$ In R, qnorm is the the
quantile function of a normal distribution. If no
mean or variance is given, then $\mu=0,\sigma=1$ are assumed.
b = qnorm(.95);  b
[1] 1.644854
diff(pnorm(c(-b,b)))
[1] 0.9

Notes: (1) Advantages of using software are that standardizing is often not necessary and rounding to use printed standard normal CDF tables can be avoided. Software sometimes gives more decimal places of accuracy than are needed in practice.
(2) When you look in the body of a printed table for
a desired probability and then get the z-score from
the margins of the table, you are doing an inverse CDF
lookup.
