# A population of values has a normal distribution with \mu=120.6 and \sigma=48.5. You intend to draw a random sample of size n=105. Find the probability that a single randomly selected value is greater than 114.9. P(X > 114.9) =? Write your answers as numbers accurate to 4 decimal places.

Random variables
A population of values has a normal distribution with $$\displaystyle\mu={120.6}$$ and $$\displaystyle\sigma={48.5}$$. You intend to draw a random sample of size $$\displaystyle{n}={105}$$.
Find the probability that a single randomly selected value is greater than 114.9.
$$\displaystyle{P}{\left({X}{>}{114.9}\right)}=$$?

2021-02-14
Step 1
Solution:
Let X be the value.
From the given information, X follows normal distribution with mean $$\displaystyle\mu={120.6}$$ and a standard deviation $$\displaystyle\sigma={48.5}$$. The sample size is 105.
Step 2
The probability that a single randomly selected value is greater than 114.9 is
$$\displaystyle{P}{\left({X}{>}{114.9}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{114.9}-{120.6}}}{{{48.5}}}}\right)}$$
$$\displaystyle={P}{\left({Z}\succ{0.118}\right)}$$
$$\displaystyle={1}-{P}{\left({Z}{<}-{0.118}\right)}$$</span>
$$\displaystyle={1}-{0.4530}={0.5470}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{0.118},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$