A population of values has a normal distribution with \mu=226.7 and \sigma=59.8. If a random sample size of n=24 is selected. Find the probability that a single randomly selected value is greater than 192.5.

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={226.7}{\quad\text{and}\quad}\sigma={59.8}$$. If a random sample size of n=24 is selected.
Find the probability that a single randomly selected value is greater than 192.5.

2021-03-03
Let $$\displaystyle{X}{f}{o}{l}{l}{o}{w}{s}{N}{\left(\mu={226.7}.\sigma={59.8}\right)}$$
The probability that a single randomly selected value is greater than 192.5 is,
$$\displaystyle{P}{\left({X}{>}{192.5}\right)}={1}-{P}{\left({X}\leq{192.5}\right)}$$
$$\displaystyle={1}-{P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\leq{\frac{{{20}-{45}}}{{{12}}}}\right)}$$
$$\displaystyle={1}-{P}{\left({z}\leq-{0.57}\right)}$$
$$\displaystyle={1}-{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left(-{0.57}\right)}\right)}$$ (Using the excel fromula)
$$\displaystyle={1}-{0.2843}={0.7157}$$
Therefore, the required probability is, 0.7157.

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