Question

The random variable X follows a normal distribution ?(20, 102).Find P(10 < ? < 35),

Random variables

The random variable X follows a normal distribution $$\displaystyle?{\left({20},{102}\right)}$$.
Find $$\displaystyle{P}{\left({10}{<}?{<}{35}\right)}$$,

2021-01-16
Step 1
Normal probability is a type of continuous probability distribution that can take random values on the whole real line. The main properties of the normal distribution are:
-It is continuous (and as a consequence, the probability of getting any single, specific outcome is zero)
-It has a "bell shaped" distribution (and that is where the "Bell-Curve" name comes along)
-The normal distribution is determined by two parameters: the population mean and population standard deviation
-It is symmetric with respect to its mean.
Given : The random variable X follows a normal distribution .
Notation: $$\displaystyle{X}\sim{N}{\left(\mu={20},\sigma^{{{2}}}={102}\right)}$$
Step 2
We need to compute $$\displaystyle{P}{r}{\left({10}\le{X}\le{35}\right)}$$.
The corresponding z-values needed to be computed are:
$$\displaystyle{Z}_{{{1}}}={\frac{{{X}_{{{1}}}-\mu}}{{\sigma}}}={\frac{{{10}-{20}}}{{{10.1}}}}=-{0.9901}$$
$$\displaystyle{Z}_{{{2}}}={\frac{{{X}_{{{2}}}-\mu}}{{\sigma}}}={\frac{{{35}-{20}}}{{{10.1}}}}={1.4851}$$
Therefore, we get:
$$\displaystyle{P}{r}{\left({10}\leq{X}\leq{35}\right)}={P}{r}{\left({\frac{{{10}-{20}}}{{{10.1}}}}\leq{Z}\leq{\frac{{{35}-{20}}}{{{10.1}}}}\right)}={P}{r}{\left(-{0.9901}\leq{Z}\leq{1.4851}\right)}$$
$$\displaystyle={P}{r}{\left({Z}\leq{1.4851}\right)}-{P}{r}{\left({Z}\leq-{0.9901}\right)}={0.9312}-{0.1611}={0.7702}$$
$$\displaystyle{P}{r}{\left({10}\leq{X}\leq{35}\right)}={0.7702}$$