## A population of values has a normal distribution with mu =67.9

GeometryA population of values has a normal distribution with mu =67.9
3 years ago
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
cleanbanditticketsdonedeal Staff answered 3 years ago
Step 1
Introduction:
Define X as the random variable of interest here.
It is given that X has a normal distribution with population mean mu = 67.9, and population standard deviation sigma = 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-(X <= 94)
=1-P ((X- mu)/ (sigma) <= (94- mu)/ (sigma))
[By standardizing]
=1-P(Z <= (94-67.9)/(44.8))
[Z= (X- mu)/ (sigma) is the standard normal variable.]
approx 1-P(Z <= 0.58)
[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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