Respuesta :
Answer:
33% probability that a man in this age group is under 180 pounds
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:
[tex]\mu = 202.3, \sigma = 50.7[/tex]
Find the probability that a man in this age group is under 180 pounds if it is known that the distribution is approximately normal.
This is the pvalue of Z when X = 180.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{180 - 202.3}{50.7}[/tex]
[tex]Z = -0.44[/tex]
[tex]Z = -0.44[/tex] has a pvalue of 0.33
33% probability that a man in this age group is under 180 pounds
