Respuesta :
Answer:
The numerical limits for a B grade are 72 and 79, that is, a score between 72 and 79 results in a B grade.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Scores on the test are normally distributed with a mean of 68.4 and a standard deviation of 9.7.
This means that [tex]\mu = 68.4, \sigma = 9.7[/tex]
Find the numerical limits for a B grade.
Below the 100 - 14 = 86th percentile and above the 65th percentile.
65th percentile:
X when Z has a p-value of 0.65, so X when Z = 0.385.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.385 = \frac{X - 68.4}{9.7}[/tex]
[tex]X - 68.4 = 0.385*9.7[/tex]
[tex]X = 72[/tex]
86th percentile:
X when Z has a p-value of 0.86, so X when Z = 1.08.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.08 = \frac{X - 68.4}{9.7}[/tex]
[tex]X - 68.4 = 1.08*9.7[/tex]
[tex]X = 79[/tex]
The numerical limits for a B grade are 72 and 79, that is, a score between 72 and 79 results in a B grade.
