Respuesta :
Answer:
The lowest score eligible for an award is 92.
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 = 82.2, \sigma = 5[/tex]
If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award
The lowest score is the 100 - 2.5 = 97.5th percentile, which is X when Z has a pvalue of 0.975. So X when Z = 1.96. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.96 = \frac{X - 82.2}{5}[/tex]
[tex]X - 82.2 = 5*1.96[/tex]
[tex]X = 92[/tex]
The lowest score eligible for an award is 92.
