# A population of values has a normal distribution with \mu = 192.3 and \sigma = 66.5. You intend to draw a random sample of size n = 15. Find the probability that a sample of size n = 15 is randomly selected with a mean less than 185.4. P(M<185.4) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={192.3}$$ and $$\displaystyle\sigma={66.5}$$. You intend to draw a random sample of size $$\displaystyle{n}={15}$$.
Find the probability that a sample of size $$\displaystyle{n}={15}$$ is randomly selected with a mean less than 185.4.
$$\displaystyle{P}{\left({M}{<}{185.4}\right)}=$$</span>?

2020-10-28
The known values are,
$$\displaystyle\mu={192.3}$$,
$$\displaystyle\sigma={66.5}$$,
$$\displaystyle{n}={15}$$
The probability that a sample size $$\displaystyle{n}={15}$$ is randomly selected with a mean less than 185.4 is,
$$\displaystyle{P}{\left({M}{<}{185.4}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{185.4}-{192.3}}}{{\frac{{66.5}}{\sqrt{{{15}}}}}}}\right)}$$</span>
$$\displaystyle={P}{\left({z}{<}-{0.402}\right)}$$</span>
$$\displaystyle={\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left(-{0.402}\right)}\right)}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {E}{x}{c}{e}{l}\ {f}{\quad\text{or}\quad}\mu{l}{a}\backslash{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({z}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}$$
$$\displaystyle={0.343842}\approx{0.3438}$$
Therefore, the required probability is, 0.3438

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