Respuesta :
Using the normal distribution, it is found that the proportion of student heights are between 94.5 centimetres and 115.5 centimetres is of 0.8664.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
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- The mean is 105, thus [tex]\mu = 105[/tex]
- The standard deviation is 7, thus [tex]\sigma = 7[/tex]
- The proportion is the p-value of Z when X = 115.5 subtracted by the p-value of Z when X = 94.5.
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X = 115.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{115.5 - 105}{7}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a p-value of 0.9332.
X = 94.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{94.5 - 105}{7}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a p-value of 0.0668.
0.9332 - 0.0668 = 0.8664.
The proportion of student heights are between 94.5 centimetres and 115.5 centimetres is of 0.8664.
A similar problem is given at https://brainly.com/question/16347292
