Respuesta :
Using the normal distribution, it is found that 3.59% of sixth-graders earn a score of at least 93 on the reading test.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- [tex]\mu[/tex] is the mean.
- [tex]\sigma[/tex] is the standard deviation.
- It measures how many standard deviations the measure is from the mean. Each z-score has a p-value associated with it.
- 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 p-value, we get the probability that the value of the measure is greater than X.
- On the reading test, the mean is of 75, thus [tex]\mu = 75[/tex].
- The standard deviation is of 10, thus [tex]\sigma = 10[/tex].
- The proportion who scored above 93 is 1 subtracted by the p-value of Z when X = 93, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{93 - 75}{10}[/tex]
[tex]Z = 1.8[/tex]
[tex]Z = 1.8[/tex] has a p-value of 0.9641.
1 - 0.9641 = 0.0359.
0.0359 x 100% = 3.59%
3.59% of sixth-graders earn a score of at least 93 on the reading test.
A similar problem is given at https://brainly.com/question/24663213
