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

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

Respuesta :

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.