Respuesta :
Answer:
a) 4,706 scored between 60 and 80
b) A score of 90.6.
c) 81th percentile.
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, we have that:
[tex]\mu = 75, \sigma = 15[/tex]
A. About how many scored between 60 and 80?
Proportion:
This is the pvalue of Z when X = 80 subtracted by the pvalue of Z when X = 60. So
X = 80
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 75}{15}[/tex]
[tex]Z = 0.33[/tex]
[tex]Z = 0.33[/tex] has a pvalue of 0.6293
X = 60
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{60 - 75}{15}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
0.6293 - 0.1587 = 0.4706
How many?
0.4706 out of 10,000, so
0.4706*10000 = 4706
4,706 scored between 60 and 80.
B. Approximately what score did only the top 15% exceed?
The 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.037 = \frac{X - 75}{15}[/tex]
[tex]X - 75 = 1.037*15[/tex]
[tex]X = 90.6[/tex]
A score of 90.6.
C. Find the percentile rank of the person who scored 88.
This is the pvalue of Z when X = 88. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{88 - 75}{15}[/tex]
[tex]Z = 0.87[/tex]
[tex]Z = 0.87[/tex] has a pvalue of 0.8078
So, rounding up, the 81th percentile.
