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3 years ago

How do I find the mean absolute deviation

Mathematics

1 answer:

hjlf3 years ago

7 0

First, you would find the mean of the data. Once you have the number, you would find the difference between the mean and the other numbers. Once you have all of those numbers, you find the mean of that set. For example:
The mean of this set is 9. We then use this number to find the difference between all of the numbers. So 14-9= 5, 9-7=2, 9-6=3, 9-5=4, 13-9=4. Finding the mean of 5, 2, 3, 4, and 4 results in 18/5 or 3.6
3.6 is the mean absolute deviation.

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3 years ago

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

Setler [38]

Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

$n=5$ ---------- computer programmers

$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

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3 years ago

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makkiz [27]

Answer:

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Step-by-step explanation:

6 is in the ones place. 3 is in the tenths place. 2 is in the hundredths place

5 is in the thousandths place.

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3 years ago

Read 2 more answers