Let $X \sim N(\mu = 16.1, \sigma^2 = 2.4^2).$ Standardize to compute $P(X > 17.8)$

Question

Let $X \sim N(u = 16.1,\, \sigma^2 = 2.4^2)$

(a) Compute $P(X > 17.8)$

(b) Compute $P(11.4 < X < 13.5)$

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No idea how to do this. IK I have to first standardize it some how, then use standard normal table.

Could someone please guide me to the right track

Answer 1

Hints:

(1) If $X\sim\mathcal N(\mu,\sigma^2)$, then $\frac{X-\mu}{\sigma}\sim\mathcal N(0,1)$. That's why you standardize -- you only have to use the standard normal distribution to compute probabilities.

(2) Since $P(X\leq c) = P(\underbrace{\frac{X-\mu}{\sigma}}_{\textrm{std normal}} \leq \frac{c-\mu}{\sigma})$, this means you just have to compute $\Phi(\frac{c-\mu}{\sigma})$.

(3) Also, $P(a<X\leq b)=P(X\leq b) - P(X\leq a)$.

(4) Finally, for continuous distributions, it doesn't matter if you have $<$ or $\leq$ at the endpoints since there is no point mass there.

Answer 2

Recall that $\frac{X - \mu }{\sigma} \sim Z \sim N(0,1)$, where $F_{Z}(z) = \Phi(z)$, hence $$ \mathbb{P}(X>17.8) = 1 - \mathbb{P}(X>17.8) = 1- \Phi\left(\frac{ 17.8 - 16.1}{2.4}\right) = 1- \Phi\left(\frac{17}{24}\right) $$ $$ \mathbb{P}(11.4 < X < 13.5 ) = \Phi\left(\frac{ 13.5 - 16.1}{2.4}\right) - \Phi\left(\frac{ 11.4 - 16.1}{2.4}\right) = \Phi\left( \frac{47}{24} \right) - \Phi\left( \frac{13}{12} \right), $$ for the last line note that $\Phi( -|a| ) = 1 - \Phi( |a| ) $

Answer 3

You have $X \sim \mathsf{Norm}(\mu = 16.1,\, \sigma=2.4)$ and you seek $P(X > 17.8).$

Using software, specifically R: Results below from R statistical software where pnorm is a normal cumulative distribution function (CDF):

1 - pnorm(17.8, 16.1, 2.4)
## 0.2393691

Standardization and printed normal table (set-up only):

$$P(X > 17.8) = P\left(\frac{X - \mu}{\sigma} > \frac{17.8 - 16.1}{2.4}\right) = P(Z > 0.708333),$$

Where $Z$ is standard normal. Using a printed table involves rounding, so you will not get exactly the same answer as from software.

(b)

diff(pnorm(c(11.4,13.5), 16.1, 2.4))
##  0.1142348