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Anna007 [38]
3 years ago
9

A die was rolled eight times. The rolls were 2, 2, 3, 4, 4, 5, 6, and 6. What is the mean absolute deviation for this data? A) 1

1/4 B) 1 2/3 C) 4 D) 10 (B is not 12/3)
Mathematics
1 answer:
adoni [48]3 years ago
6 0

Answer:

The correct option is A:

MAD = 1 + 1/4

Step-by-step explanation:

For a set of N elements {x₁, x₂, ..., xₙ}

The mean is calculated as:

M = \frac{x_1 + x_2 + ... + x_n}{N}

And the mean absolute deviation is calculated as:

MAD = \frac{Ix_1 - MI + Ix_2 - MI + ...+ IX_n - MI}{N}

Here we have the set of 8 elements:

{ 2, 2, 3, 4, 4, 5, 6, 6}

The mean of this set is:

M = (2 + 2 + 3 + 4 + 4 + 5 + 6 + 6)/8 = 4

Then the mean standard deviation is:

MAD = \frac{I2 - 4I + I2 - 4I + I3 - 4I + I4 - 4I + I4 - 4I + I5 - 4I + I6 - 4I + I6 - 4I}{8} = 10/8

If we simplify this, we get:

MAD = 10/8 = 5/4  = (4 + 1)/4 = 4/4 + 1/4 = 1 + 1/4

MAD = 1 + 1/4

The correct option is A.

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vova2212 [387]

Answer:

option (c) The mean age will stay the same but the variance will decrease

Step-by-step explanation:

Case I: For 3 executives of ages 56, 57 and 58

Number of executives, n = 3

Mean = \frac{\textup{56 + 57 + 58 }}{\textup{3}}

or

Mean = 57

Variance = \frac{\sum{(Data - Mean)^2}}{\textup{n-1}}

or

Variance = \frac{(56 - 57)^2+(57-57)^2+(58-57)^2}{\textup{3-1}}

or

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or

Variance = 1

For Case II: For 4 executives of ages 56, 57, 58 and 57

Number of executives, n = 4

Mean = \frac{\textup{56 + 57 + 58 + 57 }}{\textup{4}}

or

Mean = 57

Variance = \frac{\sum{(Data - Mean)^2}}{\textup{n-1}}

or

Variance = \frac{(56 - 57)^2+(57-57)^2+(58-57)^2+(57-57)^2}{\textup{4-1}}

or

Variance = \frac{1+0+1+0}{\textup{3}}

or

Variance = 0.67

Hence,

Mean will remain the same and the variance will decrease

Hence,

The correct answer is option (c) The mean age will stay the same but the variance will decrease

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3 years ago
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Answer:

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Step-by-step explanation:

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8x² + 16x + 3 = 0

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