A population of values has a normal distribution with \mu=73.1 and \sigma=28.1. You intend to draw a random sample of size n=131. Find the probability that a single randomly selected value is greater than 69.7. P(X > 69.7) =? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2021-02-24
A population of values has a normal distribution with $$\displaystyle\mu={73.1}$$ and $$\displaystyle\sigma={28.1}$$. You intend to draw a random sample of size $$\displaystyle{n}={131}$$.
Find the probability that a single randomly selected value is greater than 69.7.
$$\displaystyle{P}{\left({X}{>}{69.7}\right)}=$$?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-02-25
Step 1
Let us denote X= population values
Given that X follows a normal distribution with mean $$\displaystyle=\mu={73.1}$$ and standard deviation $$\displaystyle=\sigma={28.1}$$
Step 2
The probability that a single randomly selected value is greater than 69.7.
$$\displaystyle{P}{\left({X}{>}{69.7}\right)}$$
$$\displaystyle={P}{\left({Z}{>}{\frac{{{69.7}-{73.1}}}{{{28.1}}}}\right)}$$
$$\displaystyle={P}{\left({Z}\succ{0.120996441281}\right)}={0.5482}$$
Answer: 0.5482

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