Respuesta :
Answer:
Your score would be higher than 97.72% of adults, that is, higher than approximately 98% of adults.
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 = 100, \sigma = 15[/tex]
If you scored 130, your score would be higher than approximately what percent of adults?
To find the proportion of scores that are lower than, we find the pvalue of Z when X = 130. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 100}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
0.9772*100 = 97.72%.
Your score would be higher than 97.72% of adults, that is, higher than approximately 98% of adults.
