# A population of values has a normal distribution with \mu = 133.5 and \sigma = 5.2. You intend to draw a random sample of size n = 230. Find the probability that a single randomly selected value is between 133.6 and 134.1. P(133.6

Question
Random variables
A population of values has a normal distribution with $$\displaystyle\mu={133.5}$$ and $$\displaystyle\sigma={5.2}$$. You intend to draw a random sample of size $$\displaystyle{n}={230}$$.
Find the probability that a single randomly selected value is between 133.6 and 134.1.
$$\displaystyle{P}{\left({133.6}{<}{X}{<}{134.1}\right)}=$$</span>?

2021-03-19
Given Information:
$$\displaystyle\mu={133.5}$$ and $$\displaystyle\sigma={5.2}$$
$$\displaystyle{n}={230}$$
To find the probability that a single randomly selected value is between 133.6 and 134.1:
z-score is a measure which tells how many standard deviations away a data value is from the mean and is given by the formula:
$$\displaystyle{z}={\frac{{{X}-\mu}}{{\sigma}}}$$
Required probability can be obtained as follows:
$$\displaystyle{P}{\left({133.6}{<}{X}{<}{134.1}\right)}={P}{\left({X}{<}{134.1}\right)}-{P}{\left({X}{<}{133.6}\right)}$$</span>
$$\displaystyle={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{134.1}-{133.5}}}{{{5.2}}}}\right)}-{P}{\left({\frac{{{133.6}-{133.5}}}{{{5.2}}}}\right)}$$</span>
$$\displaystyle={P}{\left({Z}{<}{0.12}\right)}-{P}{\left({Z}{<}{0.02}\right)}$$</span>
$$\displaystyle={0.54776}−{0.50798}={0.03978}{\left[{u}{\sin{{g}}}{s}{\tan{{d}}}{a}{r}{d}{\left\|{a}\right\|}{l}{t}{a}{b}\le\right]}$$
Using a standard normal table, look up for z-score 0.12 i.e., 0.1 in the row and 0.02 along the column. The intersection of both values is 0.54776.
Similarly, probability corresponding to z-score 0.02 is 0.50798.
Then, subtract the lower probability value from the higher value.
Therefore, probability that a single randomly selected value is between 133.6 and 134.1 is 0.0398.

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