Respuesta :
Answer:
49.87
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.
Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%.
This means that [tex]\mu = 70, \sigma = 8[/tex]
What percentage of students receive between a 70% and 94%
The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 70. So
X = 94
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{94 - 70}{8}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a p-value of 0.9987.
X = 70
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 70}{8}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5.
0.9987 - 0.5 = 0.4987.
0.4987*100% = 49.87%.
So the percentage is 49.87%, and the answer, without the percent sign, is 49.87.
