A population of values has a normal distribution with \mu=182.5 and \sigma=49.4. 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 greater than 169.7. P(M > 169.7) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2021-03-07
A population of values has a normal distribution with \(\displaystyle\mu={182.5}\) and \(\displaystyle\sigma={49.4}\). 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 greater than 169.7.
\(\displaystyle{P}{\left({M}{>}{169.7}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-03-08
Step 1
Solution:
Let X be the value.
From the given information, X follows normal distribution with mean \(\displaystyle\mu={182.5}\) and a standard deviation \(\displaystyle\sigma={49.4}\). The sample size is 15.
Step 2
The probability that a sample of size \(\displaystyle{n}={15}\) is randomly selected with a mean greater than 169.7 is
\(\displaystyle{P}{\left({M}{>}{169.7}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{169.7}-{182.5}}}{{\frac{{49.4}}{\sqrt{{{15}}}}}}}\right)}\)
\(\displaystyle={P}{\left({Z}{>}{\frac{{\sqrt{{{15}}}{\left(-{12.8}\right)}}}{{{49.4}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\succ{1.003}\right)}\)
\(\displaystyle={1}-{0.1579}={0.8421}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{1.003},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)
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