Math Problem Statement

1. Determine the coefficient of skewness for each of the following and describe the result. a. x̄ = 60, x̃ = 50, s = 5.5 b. x̄ = 125, x̃ = 105, s = 12 c. x̄ = 250, x̃ = 295, s = 35 d. x̄ = 85, x̃ = 95, s = 7.2 2. Find the coefficient of skewness of the following data set: 5, 10, 15, 17, 20, 35. Interpret the results. 3. The mean and variance of a distribution are 40 and 625, respectively. Find the median if the skewness is -0.2.

Solution

Let's go through the three problems shown in the image:

1. Determine the coefficient of skewness for each and describe the result.

The formula for the coefficient of skewness is: Skewness=3(xˉx~)s\text{Skewness} = \frac{3(\bar{x} - \tilde{x})}{s} Where:

  • xˉ\bar{x} is the mean,
  • x~\tilde{x} is the median,
  • ss is the standard deviation.

We'll calculate the skewness for each case.

(a) xˉ=60,x~=50,s=5.5\bar{x} = 60, \tilde{x} = 50, s = 5.5

Skewness=3(6050)5.5=305.55.45\text{Skewness} = \frac{3(60 - 50)}{5.5} = \frac{30}{5.5} \approx 5.45 Since the skewness is positive, the distribution is skewed to the right.

(b) xˉ=125,x~=105,s=12\bar{x} = 125, \tilde{x} = 105, s = 12

Skewness=3(125105)12=6012=5\text{Skewness} = \frac{3(125 - 105)}{12} = \frac{60}{12} = 5 This positive skewness also indicates right-skewness.

(c) xˉ=250,x~=295,s=35\bar{x} = 250, \tilde{x} = 295, s = 35

Skewness=3(250295)35=135353.86\text{Skewness} = \frac{3(250 - 295)}{35} = \frac{-135}{35} \approx -3.86 Here, the skewness is negative, indicating left-skewness.

(d) xˉ=85,x~=95,s=7.2\bar{x} = 85, \tilde{x} = 95, s = 7.2

Skewness=3(8595)7.2=307.24.17\text{Skewness} = \frac{3(85 - 95)}{7.2} = \frac{-30}{7.2} \approx -4.17 This negative skewness also indicates left-skewness.

2. Find the coefficient of skewness of the following data set: 5, 10, 15, 17, 20, 35.

To calculate the skewness, we'll need to compute the mean, median, and standard deviation of this data set.

  • Mean (xˉ\bar{x}) = 5+10+15+17+20+356=17\frac{5 + 10 + 15 + 17 + 20 + 35}{6} = 17
  • Median (x~\tilde{x}) = 15+172=16\frac{15 + 17}{2} = 16
  • Standard deviation ss: We'll first calculate the variance.

Variance=(517)2+(1017)2+(1517)2+(1717)2+(2017)2+(3517)26=144+49+4+0+9+3246=88.33\text{Variance} = \frac{(5-17)^2 + (10-17)^2 + (15-17)^2 + (17-17)^2 + (20-17)^2 + (35-17)^2}{6} = \frac{144 + 49 + 4 + 0 + 9 + 324}{6} = 88.33 Thus, s=88.339.4s = \sqrt{88.33} \approx 9.4.

Now we can calculate the skewness: Skewness=3(1716)9.439.40.32\text{Skewness} = \frac{3(17 - 16)}{9.4} \approx \frac{3}{9.4} \approx 0.32 This is a small positive skewness, indicating a slight right-skew.

3. The mean and variance of a distribution are 40 and 625, respectively. Find the median if the skewness is -0.2.

We know:

  • Mean (xˉ\bar{x}) = 40,
  • Variance = 625, so standard deviation s=625=25s = \sqrt{625} = 25,
  • Skewness = 0.2-0.2.

Using the skewness formula: Skewness=3(xˉx~)s\text{Skewness} = \frac{3(\bar{x} - \tilde{x})}{s} Substitute the known values: 0.2=3(40x~)25-0.2 = \frac{3(40 - \tilde{x})}{25} Solve for x~\tilde{x}: 0.2×25=3(40x~)-0.2 \times 25 = 3(40 - \tilde{x}) 5=1203x~-5 = 120 - 3\tilde{x} 3x~=125x~=125341.673\tilde{x} = 125 \quad \Rightarrow \quad \tilde{x} = \frac{125}{3} \approx 41.67 So, the median is approximately 41.6741.67.


Would you like more details or explanations?

Here are 5 related questions to further explore:

  1. How does the sign of skewness affect the shape of a distribution?
  2. What are the different types of skewness and how do they impact data interpretation?
  3. How is skewness related to the mean, median, and mode of a data set?
  4. Can skewness be zero, and what does it signify?
  5. How does outlier data affect skewness?

Tip: Skewness can indicate whether data is more spread out on one side of the mean, helping to understand the distribution's shape better!

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Math Problem Analysis

Mathematical Concepts

Skewness
Descriptive Statistics
Mean
Median
Standard Deviation

Formulas

Skewness = (3 * (Mean - Median)) / Standard Deviation
Variance = sum((x - mean)^2) / n
Standard Deviation = sqrt(Variance)

Theorems

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Suitable Grade Level

Grades 10-12