Question

A population of values has a normal distribution with \mu=182.5 and \sigma=49.4. You intend to draw a random sample of size n=15. Find the probability that a single randomly selected value is greater than 169.7. P(X > 169.7) =? Write your answers as numbers accurate to 4 decimal places.

Random variables
ANSWERED
asked 2021-02-16
A population of values has a normal distribution with \(\displaystyle\mu={182.5}\) and \(\displaystyle\sigma={49.4}\). You intend to draw a random sample of size \(\displaystyle{n}={15}\).
Find the probability that a single randomly selected value is greater than 169.7.
\(\displaystyle{P}{\left({X}{>}{169.7}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-02-17

Step 1
Solution:
Let X be the value.
From the given information, X follows normal distribution with mean \(\displaystyle\mu={182.5}\) and a standard deviation \(\displaystyle\sigma={49.4}\). The sample size is 15.
Step 2
The probability that a single randomly selected value is greater than 169.7 is
\(\displaystyle{P}{\left({X}{>}{169.7}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{169.7}-{182.5}}}{{{49.4}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\succ{0.259}\right)}\)
\(\displaystyle={1}-{P}{\left({Z}{<}-{0.259}\right)}\)
\(\displaystyle={1}-{0.3978}={0.6022}\begin{bmatrix}Using\ the\ excel\ function \\=NORM.DIST(-0.259,0,1,TRUE) \end{bmatrix}\)

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