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Find the mean and mode of the distribution given the following information:

- Pearson coefficient of skewness for data distribution: 0.5
- Coefficient of variation: [tex]40\%[/tex]
- Mode: [tex]80[/tex]

Sagot :

GinnyAnswer
Sure, let's solve the problem step by step.

### Given:
- Skewness ([tex]\( g_1 \)[/tex]): 0.5
- Coefficient of Variation (CV): [tex]\( 40\% \)[/tex]
- Mode ([tex]\( M \)[/tex]): 80

First, let's convert the percentage of the coefficient of variation into a decimal form:
[tex]\[ CV = 40\% = 0.4 \][/tex]

The Pearson coefficient of skewness formula for finding the mean ([tex]\( \mu \)[/tex]) using mode ([tex]\( M \)[/tex]) is given by:
[tex]\[ \text{Skewness} = 3 \times \frac{\mu - M}{\sigma} \][/tex]
where [tex]\( \sigma \)[/tex] is the standard deviation.

Additionally, the coefficient of variation (CV) relates the standard deviation and the mean as follows:
[tex]\[ CV = \frac{\sigma}{\mu} \][/tex]

From the given data:
[tex]\[ 0.5 = 3 \times \frac{\mu - 80}{\sigma} \][/tex]
[tex]\[ CV = 0.4 = \frac{\sigma}{\mu} \][/tex]

We need to solve for the mean ([tex]\( \mu \)[/tex]). Let's first express [tex]\(\sigma\)[/tex] from the coefficient of variation formula:
[tex]\[ \sigma = 0.4 \mu \][/tex]

Substituting [tex]\(\sigma = 0.4 \mu\)[/tex] into the skewness equation:
[tex]\[ 0.5 = 3 \times \frac{\mu - 80}{0.4 \mu} \][/tex]

Rewriting this equation:
[tex]\[ 0.5 = 3 \times \frac{\mu - 80}{0.4 \mu} \][/tex]
[tex]\[ 0.5 = \frac{3 (\mu - 80)}{0.4 \mu} \][/tex]
[tex]\[ 0.5 = \frac{7.5 (\mu - 80)}{\mu} \][/tex]

Multiply both sides of the equation by [tex]\(\mu\)[/tex]:
[tex]\[ 0.5 \mu = 7.5 (\mu - 80) \][/tex]

Distribute and simplify:
[tex]\[ 0.5 \mu = 7.5 \mu - 600 \][/tex]

To isolate [tex]\(\mu\)[/tex], combine like terms by moving all terms involving [tex]\(\mu\)[/tex] to one side of the equation:
[tex]\[ 0.5 \mu - 7.5 \mu = -600 \][/tex]
[tex]\[ -7 \mu = -600 \][/tex]

Divide both sides by -7:
[tex]\[ \mu = \frac{600}{7} \][/tex]
[tex]\[ \mu \approx 85.714 \][/tex]

Thus, the mean ([tex]\( \mu \)[/tex]) of the distribution is approximately [tex]\( 85.714 \)[/tex]. The mode ([tex]\( M \)[/tex]) is given as 80.

### Final answer:
- Mean ([tex]\( \mu \)[/tex]): approximately [tex]\( 85.714 \)[/tex]
- Mode ([tex]\( M \)[/tex]): [tex]\( 80 \)[/tex]

These values satisfy the conditions provided in the problem statement.
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