A population of values has a normal distribution with mu =67.9 and sigma =44.8. You intend to draw a random sample of size n=17. Find the probability that a single randomly selected value is greater than 94

Eisregiony1

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2022-08-18

A population of values has a normal distribution with μ=67.9 and σ=44.8. You intend to draw a random sample of size n=17.
Find the probability that a single randomly selected value is greater than 94

Answer & Explanation

Andre Ferguson

Andre Ferguson

Beginner2022-08-19Added 12 answers

Step 1
Introduction:
Define X as the random variable of interest here.
It is given that X has a normal distribution with population mean μ=67.9, and population standard deviation σ=44.8.
A sample of size n = 17 is selected from the population.
Step 2
Calculation:
The probability that a single randomly selected value is greater than 94 is calculated below:
P(X>94)=1(X94)
=1P(Xμσ94μσ)
[By standardizing]
=1P(Z9467.944.8)
[Z=Xμσ is the standard normal variable.]
1P(Z0.58)
[The z-score is rounded to two decimal places.]
=1-0.7190
[Using the standard normal distribution table]
=0.2810
Thus, the probability that a single randomly selected value is greater than 94 is 0.2810.

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