College

In human engineering and product design, it is important to consider the weights of people so that airplanes or elevators are not overloaded. Based on data from the National Health Survey, the weight for adult males in the U.S. follows a bell-shaped distribution with a mean weight of 173 pounds and a standard deviation of 30 pounds. Using the z-score approach for detecting outliers, which of the following weights would represent potential outliers in the distribution of U.S. adult male weights?

A. 110 pounds
B. 157 pounds
C. 281 pounds

Answer :

Answer:

A. 110 pounds,

C. 281 pounds

Step-by-step explanation:

Problems of normally distributed samples can be solved using 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.

A measure is said to be an outlier if it has a pvalue lesser than 0.05 or higher than 0.95.

In this problem, we have that:

[tex]\mu = 173, \sigma = 30[/tex]

A. 110 pounds,

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{110 - 173}{30}[/tex]

[tex]Z = -2.1[/tex]

[tex]Z = -2.1[/tex] has a pvalue of 0.0179. So a weight of 110 pounds is an outlier.

B. 157 pounds,

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{157 - 173}{30}[/tex]

[tex]Z = 0.53[/tex]

[tex]Z = 0.53[/tex] has a pvalue of 0.702.

So a weight of 157 is not an outlier.

C. 281 pounds

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{281 - 173}{30}[/tex]

[tex]Z = 3.6[/tex]

[tex]Z = 3.6[/tex] has a pvalue of 0.9988.

So a weight of 281 is an outlier.

Final answer:

In order to determine outliers based on z-score calculation, only weight C (281 pounds) is considered as an outlier since it lies more than 3 standard deviations from the mean. Weights A and B are within 3 standard deviations, so they're not considered outliers.

Explanation:

In statisticians' terms, an outlier is typically considered to be any data point more than 1.5 imes the interquartile range (IQR) above the third quartile or below the first quartile. However, in this case, since we are using the z-score approach, we consider any data point that lies more than 3 standard deviations from the mean as being significantly different – or an outlier.

The z-score is a measure of how many standard deviations an element is from the mean. To find out if any of these weights are outliers, we need to convert each weight into a z-score. The z-score is calculated by the formula Z = (X - μ)/σ where X is the data point, μ is the mean and σ is the standard deviation. In this case, the mean μ = 173 pounds and the standard deviation σ = 30 pounds.

  • For weight A (110 pounds), its z-score would be Z = (110-173)/30 = -2.1
  • For weight B (157 pounds), its z-score would be Z = (157-173)/30 = -0.53
  • For weight C (281 pounds), its z-score would be Z = (281-173)/30 = 3.6

So, in the context of this problem, considering a threshold of 3 standard deviations, only weight C (281 pounds) would be considered as an outlier.

Learn more about Outliers here:

https://brainly.com/question/34952867

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