A random variable $X$ is normally distributed with mean $µ$ and standard deviation $σ$.
lf $P(X < 25)$ is $0.1082$ and $P(X> 42)$ is $0.1303$ determine:
(i) $P(25 < X < 42)$
(ii) the values of mean $µ$ and standard deviation $σ$
(iii) $P(X>22)$
For part (i), I have done $P(25 < X < 42) = P(X<42) – P(X<25)$ which gives $(1-0.1303)-0.1082 = 0.7615$.
I'm unsure how to do part (ii). I'm thinking it has something to do with the empirical rule, so $68.3\%$ lies within one standard deviation of the mean, etc. But I'm not sure how to apply this, as there doesn't seem to be any concrete "percent rule" that the $0.7615$ applies to. This is from a high school maths textbook. Any hints would be appreciated.
Best Answer
In normal distribution we use the notation $P(X\le a)=\phi(a)$
$P\bigg(\dfrac{x-\mu}{\sigma} < \dfrac {25-\mu}{\sigma}\bigg)=P\bigg(Z< \dfrac {25-\mu}{\sigma}\bigg)=\phi\bigg(\dfrac{25-\mu}{\sigma}\bigg)=0.1082$
Now look at normal table and find $a$ such that $\phi(a)=0.1082$ then,
$a=\dfrac{25-\mu}{\sigma}$
similarly, you have $P(X>42)=.1303$
So you will get two equations two unknows$(\mu$ and $\sigma)$ solve them you will get $ \mu,\sigma$