Respuesta :
Answer:
a) 88.10% of students would be expected to score over 80
b) 25.78% of students would be expected to score under 89.
c) 22.34% of students would be expected to score between 111 and 131.
d) 4.95% of students would be expected to score over 128.
Step-by-step explanation:
Problems of normally distributed samples are 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 problem, we have that:
[tex]\mu = 100, \sigma = 17[/tex]
a) over 80?
1 subtracted by the pvalue of Z when X = 80.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 100}{17}[/tex]
[tex]Z = -1.18[/tex]
[tex]Z = -1.18[/tex] has a pvalue of 0.1190
1 - 0.1190 = 0.8810
88.10% of students would be expected to score over 80
b) under 89?
Pvalue of Z when X = 89. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 100}{17}[/tex]
[tex]Z = -0.65[/tex]
[tex]Z = -0.65[/tex] has a pvalue of 0.2578.
25.78% of students would be expected to score under 89.
c) between 111 and 131?
Pvalue of Z when X = 131 subtracted by the pvalue of Z when X = 111.
X = 131
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{131 - 100}{17}[/tex]
[tex]Z = 1.82[/tex]
[tex]Z = 1.82[/tex] has a pvalue of 0.9656.
X = 111
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{111 - 100}{17}[/tex]
[tex]Z = 0.65[/tex]
[tex]Z = 0.65[/tex] has a pvalue of 0.7422.
0.9656 - 0.7422 = 0.2234
22.34% of students would be expected to score between 111 and 131.
d) over 128?
1 subtracted by the pvalue of Z when X = 128. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{128 - 100}{17}[/tex]
[tex]Z = 1.65[/tex]
[tex]Z = 1.65[/tex] has a pvalue of 0.9505
1 - 0.9505 = 0.0495
4.95% of students would be expected to score over 128.
