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hoa [83]
3 years ago
9

The mean is a better descriptor for skewed distribution. a. true b. false

Mathematics
2 answers:
Molodets [167]3 years ago
7 0

The <em><u>correct answer</u></em> is:

b. false

Explanation:

When a distribution is skewed, this means that there is at least one value that changes the shape of the distribution.  This value is much different from the rest of the data; this makes it an outlier.

An outlier has an affect on the mean; if it is higher than the rest of the data, it increases the mean.  If the outlier is lower than the rest of the data, it decreases the mean.

vitfil [10]3 years ago
4 0
Most likely true. It depends. When the distribution is skewed with an extreme tilting it to one side, median would be a better measure of central tendency. If the extremities are almost balanced on both sides then mean would be a better measure of central tendency.
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Answer:

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Step-by-step explanation:

Given that:

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The probability that the mean student loan debt for these people is between $31000 and $33000 can be computed as:

P(31000 < X < 33000) = P( X \leq 33000) - P (X \leq 31000)

P(31000 < X < 33000) = P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P(Z \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{3000}{\dfrac{9000}{10}}}) -P(Z \leq \dfrac{1000}{\dfrac{9000}{10}}})

P(31000 < X < 33000) = P( Z \leq 3.33)-P(Z \leq 1.11})

From Z tables:

P(31000 < X

P(31000 < X

Therefore; the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

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