Statistics Examples

10

$10$ ,

74

$74$ ,

78

$78$ ,

102

$102$

Step 1

The mean of a set of numbers is the sum divided by the number of terms.

μ = \frac{10 + 74 + 78 + 102}{4}

$μ = \frac{10 + 74 + 78 + 102}{4}$

Step 2

Cancel the common factor of

10 + 74 + 78 + 102

$10 + 74 + 78 + 102$ and

4

$4$ .

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Step 2.1

Factor

2

$2$ out of

10

$10$ .

μ = \frac{2 (5) + 74 + 78 + 102}{4}

$μ = \frac{2 (5) + 74 + 78 + 102}{4}$

Step 2.2

Factor

2

$2$ out of

74

$74$ .

μ = \frac{2 \cdot 5 + 2 \cdot 37 + 78 + 102}{4}

$μ = \frac{2 \cdot 5 + 2 \cdot 37 + 78 + 102}{4}$

Step 2.3

Factor

2

$2$ out of

2 \cdot 5 + 2 \cdot 37

$2 \cdot 5 + 2 \cdot 37$ .

μ = \frac{2 \cdot (5 + 37) + 78 + 102}{4}

$μ = \frac{2 \cdot (5 + 37) + 78 + 102}{4}$

Step 2.4

Factor

2

$2$ out of

78

$78$ .

μ = \frac{2 \cdot (5 + 37) + 2 (39) + 102}{4}

$μ = \frac{2 \cdot (5 + 37) + 2 (39) + 102}{4}$

Step 2.5

Factor

2

$2$ out of

2 \cdot (5 + 37) + 2 (39)

$2 \cdot (5 + 37) + 2 (39)$ .

μ = \frac{2 \cdot (5 + 37 + 39) + 102}{4}

$μ = \frac{2 \cdot (5 + 37 + 39) + 102}{4}$

Step 2.6

Factor

2

$2$ out of

102

$102$ .

μ = \frac{2 \cdot (5 + 37 + 39) + 2 (51)}{4}

$μ = \frac{2 \cdot (5 + 37 + 39) + 2 (51)}{4}$

Step 2.7

Factor

2

$2$ out of

2 \cdot (5 + 37 + 39) + 2 (51)

$2 \cdot (5 + 37 + 39) + 2 (51)$ .

μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{4}

$μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{4}$

Step 2.8

Cancel the common factors.

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Step 2.8.1

Factor

2

$2$ out of

4

$4$ .

μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{2 (2)}

$μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{2 (2)}$

Step 2.8.2

Cancel the common factor.

μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{2 \cdot 2}

$μ = \frac{2 \cdot (5 + 37 + 39 + 51)}{2 \cdot 2}$

Step 2.8.3

Rewrite the expression.

μ = \frac{5 + 37 + 39 + 51}{2}

$μ = \frac{5 + 37 + 39 + 51}{2}$

μ = \frac{5 + 37 + 39 + 51}{2}

$μ = \frac{5 + 37 + 39 + 51}{2}$

μ = \frac{5 + 37 + 39 + 51}{2}

$μ = \frac{5 + 37 + 39 + 51}{2}$

Step 3

Simplify the numerator.

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Step 3.1

Add

5

$5$ and

37

$37$ .

μ = \frac{42 + 39 + 51}{2}

$μ = \frac{42 + 39 + 51}{2}$

Step 3.2

Add

42

$42$ and

39

$39$ .

μ = \frac{81 + 51}{2}

$μ = \frac{81 + 51}{2}$

Step 3.3

Add

81

$81$ and

51

$51$ .

μ = \frac{132}{2}

$μ = \frac{132}{2}$

μ = \frac{132}{2}

$μ = \frac{132}{2}$

Step 4

Divide

132

$132$ by

2

$2$ .

μ = 66

$μ = 66$

Step 5

Arrange the terms in ascending order.

M e d i a n = 10, 74, 78, 102

$M e d i a n = 10, 74, 78, 102$

Step 6

The median is the middle term in the arranged data set. In the case of an even number of terms, the median is the average of the two middle terms.

M e d i a n = \frac{74 + 78}{2}

$M e d i a n = \frac{74 + 78}{2}$

Step 7

Remove parentheses.

M e d i a n = \frac{74 + 78}{2}

$M e d i a n = \frac{74 + 78}{2}$

Step 8

Cancel the common factor of

74 + 78

$74 + 78$ and

2

$2$ .

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Step 8.1

Factor

2

$2$ out of

74

$74$ .

M e d i a n = \frac{2 \cdot 37 + 78}{2}

$M e d i a n = \frac{2 \cdot 37 + 78}{2}$

Step 8.2

Factor

2

$2$ out of

78

$78$ .

M e d i a n = \frac{2 \cdot 37 + 2 \cdot 39}{2}

$M e d i a n = \frac{2 \cdot 37 + 2 \cdot 39}{2}$

Step 8.3

Factor

2

$2$ out of

2 \cdot 37 + 2 \cdot 39

$2 \cdot 37 + 2 \cdot 39$ .

M e d i a n = \frac{2 \cdot (37 + 39)}{2}

$M e d i a n = \frac{2 \cdot (37 + 39)}{2}$

Step 8.4

Cancel the common factors.

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Step 8.4.1

Factor

2

$2$ out of

2

$2$ .

M e d i a n = \frac{2 \cdot (37 + 39)}{2 (1)}

$M e d i a n = \frac{2 \cdot (37 + 39)}{2 (1)}$

Step 8.4.2

Cancel the common factor.

M e d i a n = \frac{2 \cdot (37 + 39)}{2 \cdot 1}

$M e d i a n = \frac{2 \cdot (37 + 39)}{2 \cdot 1}$

Step 8.4.3

Rewrite the expression.

M e d i a n = \frac{37 + 39}{1}

$M e d i a n = \frac{37 + 39}{1}$

Step 8.4.4

Divide

37 + 39

$37 + 39$ by

1

$1$ .

M e d i a n = 37 + 39

$M e d i a n = 37 + 39$

M e d i a n = 37 + 39

$M e d i a n = 37 + 39$

M e d i a n = 37 + 39

$M e d i a n = 37 + 39$

Step 9

Add

37

$37$ and

39

$39$ .

M e d i a n = 76

$M e d i a n = 76$

Step 10

Convert the median

76

$76$ to decimal.

M e d i a n = 76

$M e d i a n = 76$

Step 11

Since the mean is less than the median, the data set is negatively skewed.

Negatively Skewed

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