# A population of values has a normal distribution with \mu=197 and \sigma=68. You intend to draw a random sample of size n=181 Find the probability that a sample of size n=181 is randomly selected with a mean between 198.5 and 205.6. P(198.5 < M < 205.6) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={197}$$ and $$\displaystyle\sigma={68}$$. You intend to draw a random sample of size $$\displaystyle{n}={181}$$
Find the probability that a sample of size $$\displaystyle{n}={181}$$ is randomly selected with a mean between 198.5 and 205.6.
$$\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}=$$</span>?

2021-01-29
Step 1
From the provided information,
Mean $$\displaystyle{\left(\mu\right)}={197}$$
Standard deviation $$\displaystyle{\left(\sigma\right)}={68}$$
Let X be a random variable which represents the value.
$$\displaystyle{X}\sim{N}{\left({197},{68}\right)}$$
Step 2
Sample size $$\displaystyle{\left({n}\right)}={181}$$
The required probability that a sample of size $$\displaystyle{n}={181}$$ is randomly selected with a mean between 198.5 and 205.6 $$\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}$$</span> can be obtained as:
$$\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}={P}{\left({\frac{{{198.5}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{205.6}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}\right)}$$</span>
$$\displaystyle={P}{\left({0.297}{<}{Z}{<}{1.701}\right)}$$</span>
$$\displaystyle={P}{\left({Z}{<}{1.701}\right)}-{P}{\left({Z}{<}{0.297}\right)}$$</span>
$$\displaystyle={0.9555}-{0.6168}={0.3387}$$ (Using standard normal table)
Thus, the required probability is 0.3387.

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