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# A population of values has a normal distribution with \mu = 181 and \sigma = 41.8. You intend to draw a random sample of size n = 144. Find the probability that a single randomly selected value is between 176.5 and 183.1. P(176.5<X<183.1)=? Write your answers as numbers accurate to 4 decimal places.

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Random variables
asked 2021-02-05
A population of values has a normal distribution with $$\displaystyle\mu={181}$$ and $$\displaystyle\sigma={41.8}$$. You intend to draw a random sample of size $$\displaystyle{n}={144}$$.
Find the probability that a single randomly selected value is between 176.5 and 183.1.
$$\displaystyle{P}{\left({176.5}{<}{X}{<}{183.1}\right)}=$$</span>?
Write your answers as numbers accurate to 4 decimal places.

## Answers (1)

2021-02-06
Given that,
$$\displaystyle\mu={181}$$,
$$\displaystyle\sigma={41.8}$$
$$\displaystyle{n}={144}$$
The probability that a single randomly selected value is between 176.5 and 183.1 is,
$$\displaystyle{P}{\left({176.5}{<}{X}{<}{183.1}\right)}={P}{\left({\frac{{{176.5}-{181}}}{{{41.8}}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{183.1}-{181}}}{{{41.8}}}}\right)}$$</span>
$$\displaystyle={P}{\left(-{0.108}{<}{z}{<}{0.050}\right)}$$</span>
$$\displaystyle={P}{\left({z}{<}{0.050}\right)}-{P}{\left({z}{<}-{0.108}\right)}$$</span>
$$\displaystyle=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({0.050}\right)}\right)}\backslash-{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left(-{0.108}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left({U}{\sin{{g}}}\ {E}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n},\backslash{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({z}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}\right.}$$
$$\displaystyle={0.5199}-{0.4570}={0.0629}$$
Therefore, the required probability is, 0.0629.

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