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 single randomly selected value is greater than 72.4. P(X>72.4)=? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2020-11-02
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 single randomly selected value is greater than 72.4.
\(\displaystyle{P}{\left({X}{>}{72.4}\right)}=\)?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2020-11-03
Step 1
Given,
Mean = 49
Standard deviation = 79.5
Sample size = 84
Step 2
Consider,
\(\displaystyle{P}{\left({X}{>}{72.4}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{72.4}-\mu}}{{\sigma}}}\right)}\)
\(\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-{49}}}{{{79.5}}}}\right)}\)
\(\displaystyle={P}{\left({Z}{>}{0.294}\right)}\)
\(\displaystyle={1}-{P}{\left({Z}\leq{0.294}\right)}\)
\(\displaystyle={1}-{0.6156}={0.3844}\) (From the standard normal table)
The probability that a single randomly selected value is greater than 72.4 is, 0.3844.
0

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