Question

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introverts. (For each answer, enter a number. Round your answers to three decimal places.)

Answer 1

`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

Explanation:

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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`