Using the normal distribution, it is found that 0.70% of the measures are considered outliers using the 1.5IQR rule.
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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 IQR is the difference between the 75th and the 25th percentile.
- The 25th percentile is X when Z has a p-value of 0.25, so X when Z = -0.675.
- The 75th percentile is X when Z has a p-value of 0.75, so X when Z = 0.675.
- Finding the z-scores outside the 1.5IQR range, we get that they are:
[tex]Z_l = -0.675 - 1.5(|-0.675-0.675|) = -2.7[/tex]
[tex]Z_h = 0.675 + 1.5(|-0.675-0.675|) = 2.7[/tex]
- Z = 2.7 has a p-value of 0.9965, 1 - 0.9965 = 0.0035.
- Considering the values below Z = -2.7, also 0.0035 proportion.
- [tex]2 \times 0.0035 = 0.0070, 0.007 \times 100\% = 0.7\%[/tex]
- Thus, 0.70% of the measures are considered outliers using the 1.5IQR rule.
A similar problem is given at https://brainly.com/question/14243195
