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# A population of values has a normal distribution with \mu = 49 and \sigma = 79.5. You intend to draw a random sample of size n=84. Find the probability that a a sample of size n = 84 is randomly selected with a mean greater than 72.4. P(M>72.4)=? Write your answers as numbers accurate to 4 decimal places.

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Random variables
asked 2020-10-27
A population of values has a normal distribution with $$\displaystyle\mu={49}$$ and $$\displaystyle\sigma={79.5}$$. You intend to draw a random sample of size $$\displaystyle{n}={84}$$.
Find the probability that a a sample of size $$\displaystyle{n}={84}$$ is randomly selected with a mean greater than 72.4.
$$\displaystyle{P}{\left({M}{>}{72.4}\right)}=$$?
Write your answers as numbers accurate to 4 decimal places.

## Answers (1)

2020-10-28
Step 1
Given,
Mean = 49
Standard deviation = 79.5
Sample size = 84
Step 2
Consider,
$$\displaystyle{P}{\left({M}{>}{72.4}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}$$
$$\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}$$
$$\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-{49}}}{{\frac{{79.5}}{\sqrt{{{84}}}}}}}\right)}$$
$$\displaystyle={P}{\left({Z}{>}{2.70}\right)}$$
$$\displaystyle={1}-{P}{\left({Z}\leq{2.70}\right)}$$
$$\displaystyle={1}-{0.9965}={0.0035}$$ (Using standard normal table)
The probability that a sample of size $$\displaystyle{n}={84}$$ is randomly selected with a mean greater than 72.4 is, 0.0035.

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