Respuesta :
Answer:
27% of the possible Z values are greater than 0.613
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 = 0, \sigma = 1[/tex]
27% of the possible Z values are greater than
The 100 - 27 = 73rd percentile, which is X when Z has a pvalue of 0.73. So X when the z-score is 0.613.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.613 = \frac{X - 0}{1}[/tex]
[tex]X = 0.613[/tex]
27% of the possible Z values are greater than 0.613
