Is a measure of 26 inches​ "far away" from a mean of 16 ​inches? As someone with knowledge of​ statistics, you answer​ "it depends" and request the standard deviation of the underlying data. ​(a) Suppose the data come from a sample whose standard deviation is 2 inches. How many standard deviations is 26 inches from 16 ​inches? ​(b) Is 26 inches far away from a mean of 16 ​inches? ​(c) Suppose the standard deviation of the underlying data is 7 inches. Is 26 inches far away from a mean of 16 ​inches?

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Answer: Hello!

In a normal distribution, between the mean and the mean plus the standar deviation, there is a 34.1% of the data set, between the mean plus the standar deviation, and the mean between two times the standard deviation, there is a 16.2% of the data set, and so on.

If our mean is 16 inches, and the measure is 26 inches, then the difference is 10 inches between them.

a) if the standar deviation is 2 inches, then you are 10/2 = 5 standar deviations from the mean.

b) yes, is really far away from the mean, in a normal distribution a displacement of 5 standar deviations has a very small probability.

c) Now the standar deviation is 7, so now 26 is in the range between 1 standar deviation and 2 standar deviations away from the mean.

Then this you have a 16% of the data, then in this case, 26 inches is not far away from the mean.

Using z-scores, it is found that:

  • a) 26 inches is 5 standard deviations from 16 inches.
  • b) |Z| = 5 > 2, thus yes, 26 inches is far away from a mean of 16 ​inches.
  • c) |Z| = 1.43 < 2, thus, 26 inches is not far away from a mean of 16 ​inches.

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  • 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.
  • If |Z| > 2, the measure is said to be far from the mean.

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  • Mean of 16 means that [tex]\mu = 16[/tex].
  • Measure of 26 means that [tex]X = 26[/tex].

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Item a:

  • Standard deviation of 2, thus [tex]\sigma = 2[/tex].

The z-score is:

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

[tex]Z = \frac{26 - 16}{2}[/tex]

[tex]Z = 5[/tex]

Thus, 26 inches is 5 standard deviations from 16 inches.

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Item b:

|Z| = 5 > 2, thus yes, 26 inches is far away from a mean of 16 ​inches.

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Item c:

  • Standard deviation of 7, thus, [tex]\sigma = 7[/tex]

The z-score is:

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

[tex]Z = \frac{26 - 16}{7}[/tex]

[tex]Z = 1.43[/tex]

|Z| = 1.43 < 2, thus, 26 inches is not far away from a mean of 16 ​inches.

A similar problem is given at https://brainly.com/question/24126815

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