Respuesta :
Answer:
The range that you would expect 68.26 percent of the grades to fall is between 53 and 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]
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 = 68, \sigma = 15[/tex]
Middle 68.26% of the grades:
From the
50 - (68.26/2) = 15.87th percentile
To the
50 + (68.26/2) = 84.13rd percentile.
15.87th percentile:
X when Z has a pvalue of 0.1587. So X when Z = -1.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1 = \frac{X - 68}{15}[/tex]
[tex]X - 68 = -1*15[/tex]
[tex]X = 53[/tex]
84.13rd percentile:
X when Z has a pvalue of 0.8413. So X when Z = 1.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1 = \frac{X - 68}{15}[/tex]
[tex]X - 68 = 1*15[/tex]
[tex]X = 83[/tex]
The range that you would expect 68.26 percent of the grades to fall is between 53 and 83.
