Respuesta :
Answer:
a) 0.84% of students scored over 32
b) 29.12% of students scored under 17
c) 70.04% of students scored between 17 and 32.
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 = 19.8, \sigma = 5.1[/tex]
a) About what percent of students scored over 32?
This is 1 subtracted by the pvalue of Z when X = 32. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 19.8}{5.1}[/tex]
[tex]Z = 2.39[/tex]
[tex]Z = 2.39[/tex] has a pvalue of 0.9916
1 - 0.9916 = 0.0084
So 0.84% of students scored over 32
b) About what percent of students scored under 17?
This is the pvalue of Z when X = 17. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{17 - 19.8}{5.1}[/tex]
[tex]Z = -0.55[/tex]
[tex]Z = -0.55[/tex] has a pvalue of 0.2912
So 29.12% of students scored under 17
c) About what percent of students scored between 17 and 32?
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 12. So
X = 32
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 19.8}{5.1}[/tex]
[tex]Z = 2.39[/tex]
[tex]Z = 2.39[/tex] has a pvalue of 0.9916
X = 17
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{17 - 19.8}{5.1}[/tex]
[tex]Z = -0.55[/tex]
[tex]Z = -0.55[/tex] has a pvalue of 0.2912
0.9916 - 0.2912 = 0.7004
70.04% of students scored between 17 and 32.
About 70 percent of students scored between 17 and 32
Z score
The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (x - μ) / σ
where μ is the mean, x = raw score and σ is the standard deviation.
a) Given μ = 19.8, σ = 5.1. For x > 32:
z = (32 - 19.8)/5.1 = 2.39
P(z > 2.39) = 1 - P(z < 2.39) = 1 - 0.9916 = 0.84%
b) For x < 17:
z = (17 - 19.8)/5.1 = 0.2912 = 29.12%
c) P(17 < x < 32) = P(-0.55 < z < 2.39) = P(z < 2.39) - P(z < -0.55) = 0.9916 - 0.2912 = 70%
About 70 percent of students scored between 17 and 32
Find out more on Z score at: https://brainly.com/question/25638875
