In the data set below, what is the mean absolute deviation? 3 8 6 9 2 5

Mathematics

1 answer:

zepelin [54]3 years ago

6 0

<h3>Answer: Approximately 2.1667</h3>

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Explanation:

First we need to find the arithmetic mean.

Add up the values to get 3+8+6+9+2+5 = 33

Divide this over 6 because there are 6 values: 33/6 = 5.5

The mean is 5.5

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Subtract the mean from each data value. Apply absolute value to ensure the result is not negative.

|3 - 5.5| = |-2.5| = 2.5
|8 - 5.5| = |2.5| = 2.5
|6 - 5.5| = |0.5| = 0.5
|9 - 5.5| = |3.5| = 3.5
|2 - 5.5| = |-3.5| = 3.5
|5 - 5.5| = |-0.5| = 0.5

Each result represents how far that specific data value is from the mean. Absolute value is used to find distance on a number line.

Add up each of those results to get: 2.5+2.5+0.5+3.5+3.5+0.5 = 13

Then divide by 6 to find the average: 13/6 = 2.1667 which is approximate

The mean absolute deviation represents the average distance any given value is from the mean. The term "deviation" refers to how far each data value is from the mean, which is the center point we're after more or less.

The MAD (mean absolute deviation) effectively measures how spread out the data set is. The larger the MAD, the more spread out the data is likely to be. The range, standard deviation and variance are also measures of spread.

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Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$