Respuesta :
Answer:
26.11% of the test scores during the past year exceeded 83.
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]
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:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that [tex]\mu = 78, \sigma = 7.8[/tex].
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when [tex]X = 83[/tex]. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 78}{7.8}[/tex]
[tex]Z = 0.64[/tex]
[tex]Z = 0.64[/tex] has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.
