Respuesta :
Answer:
25.14% probability that his score is at least 582.5.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 506, \sigma = 114[/tex]
If 1 of the men is randomly selected, find the probability that his score is at least 582.5.
This is 1 subtracted by the pvalue of Z when X = 582.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{582.5 - 506}{114}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a pvalue of 0.7486
1 - 0.7486 = 0.2514
25.14% probability that his score is at least 582.5.
