Question

A population of values has a normal distribution with mean =136.4 and standard deviation =30.2. A random sample of size n=158 is drawn. Find the probability that a single randomly selected value is greater than 135. Roung your answer to four decimal places. P(X>135)=?

Random variables
ANSWERED
asked 2021-01-10
A population of values has a normal distribution with mean =136.4 and standard deviation =30.2. A random sample of size \(\displaystyle{n}={158}\) is drawn.
Find the probability that a single randomly selected value is greater than 135. Roung your answer to four decimal places.
\(\displaystyle{P}{\left({X}{>}{135}\right)}=\)?

Answers (1)

2021-01-11

Step 1
The Z-score of a random variable X is defined as follows:
\(\displaystyle{Z}=\frac{{{X}-\mu}}{\sigma}\)
Here, \(\displaystyle\mu{\quad\text{and}\quad}\sigma\) are the mean and standard deviation of X, respectively.
Step 2
Consider a random variable X, that defines the variable of interest.
According to the given information, X follows normal distribution with mean \(\displaystyle\mu_{{{x}}}={136.4}\) and the standard deviation of \(\displaystyle\sigma_{{{x}}}={30.2}\).
The probability that a single randomly selected value is greater than 135 is,
\(\displaystyle{P}{\left({X}{>}{135}\right)}={1}-{P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\leq{\frac{{{135}-{136.4}}}{{{30.2}}}}\right)}\)
\(\displaystyle={1}-{P}{\left({Z}\leq-{0.046357615}\right)}\)
\(=1-0.4815 \begin{bmatrix}using\ the\ Excel\ formula\\ =NORM.S.DIST(-0.046357615,TRUE) \end{bmatrix} =0.5185\)
Therefore, the probability that a single randomly selected value is greater than 135 is 0.5185.

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