Respuesta :
Answer:
0.7422 = 74.22% of scores are above 74.
Step-by-step explanation:
Problems of normally distributed samples can be solved using 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 problem, we have that:
[tex]\mu = 78, \sigma = 6.11[/tex]
Calculate the proportion of scores above 74.
This is 1 subtracted by the pvalue of Z when X = 74. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{74 - 78}{6.11}[/tex]
[tex]Z = -0.65[/tex]
[tex]Z = -0.65[/tex] has a pvalue of 0.2578
So 1-0.2578 = 0.7422 = 74.22% of scores are above 74.
