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Ad libitum [116K]
2 years ago
14

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

s. (For each answer, enter a number. Round your answers to three decimal places.)
Mathematics
1 answer:
Setler [38]2 years ago
3 0

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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