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Sagot :
Answer:
The proportion of the population used to develop the test that has scores below 1.7 is 0.063.
The proportion of the population used to develop the test that has scores between 1.7 and 2.1 is 0.0662.
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.
The distribution of ARSMA scores in a population used to develop the test is approximately Normal with mean 3.0 and standard deviation 0.8.
This means that [tex]\mu = 3, \sigma = 0.8[/tex]
What proportion of the population used to develop the test has scores below 1.7?
This is the pvalue of Z when X = 1.7. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.7 - 3}{0.8}[/tex]
[tex]Z = -1.63[/tex]
[tex]Z = -1.63[/tex] has a pvalue of 0.063
The proportion of the population used to develop the test that has scores below 1.7 is 0.063.
Between 1.7 and 2.1?
This is the pvalue of Z when X = 2.1 subtracted by the pvalue of Z when X = 1.7.
X = 2.1
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.1 - 3}{0.8}[/tex]
[tex]Z = -1.13[/tex]
[tex]Z = -1.13[/tex] has a pvalue of 0.1292
X = 1.7
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.7 - 3}{0.8}[/tex]
[tex]Z = -1.63[/tex]
[tex]Z = -1.63[/tex] has a pvalue of 0.063
0.1292 - 0.063 = 0.0662
The proportion of the population used to develop the test that has scores between 1.7 and 2.1 is 0.0662.
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