# 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 sample of size n=105 is randomly selected with a mean greater than 114.9. P(M > 114.9) =? Write your answers as numbers accurate to 4 decimal places. Question
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 sample of size $$\displaystyle{n}={105}$$ is randomly selected with a mean greater than 114.9.
$$\displaystyle{P}{\left({M}{>}{114.9}\right)}=$$? 2020-10-26
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 sample of size $$\displaystyle{n}={105}$$ is randomly selected with a mean greater than 114.9 is
$$\displaystyle{P}{\left({M}{>}{114.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{114.9}-{120.6}}}{{\frac{{48.5}}{\sqrt{{{105}}}}}}}\right)}$$
$$\displaystyle={P}{\left({Z}{>}{\frac{{\sqrt{{{105}}}{\left(-{5.7}\right)}}}{{{48.5}}}}\right)}$$
$$\displaystyle={P}{\left({Z}\succ{1.204}\right)}$$
$$\displaystyle={1}-{P}{\left({Z}{<}-{1.204}\right)}$$</span>
$$\displaystyle={1}-{0.1143}={0.8857}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{0.118},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$

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