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Sagot :
Using the normal distribution, it is found that there is a 62.4% probability that a data value is between 37 and 41.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 38, \sigma = 2[/tex]
As a proportion, the probability that a data value is between 37 and 41 is the p-value of Z when X = 41 subtracted by the p-value of Z when X = 37, hence:
X = 41:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{41 - 38}{2}[/tex]
Z = 1.5
Z = 1.5 has a p-value of 0.933.
X = 37:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{37 - 38}{2}[/tex]
Z = -0.5
Z = -0.5 has a p-value of 0.309.
0.933 - 0.309 = 0.624,
0.624 = 62.4% probability that a data value is between 37 and 41.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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