Question

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are introverts. (Round your answers to three decimal places.) (a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

Answer 1

Step-by-step explanation:

Let 'X' be a random variable representing the number of extroverts.

Given approximately 80% of all marketing personnel are extroverts, hence, probability of success will be

p=0.80andq=1−0.80=0.2.

Given, in a meeting there are 15 marketing personnel present.

Then

X∼Bin(15,0.8)

(a) at least 10 are extroverts.

P(X≥10)=P(X=10)+P(X=11)+P(X=12)+P(X=13)P(X=14)+P(X=15)=15C10(0.8)10(0.2)5+15C15(0.8)15(0.2)0=0.939

At least 5 are extroverts.

P(X≥5)=1−P(X=0)−P(X=1)−P(X=2)−P(X=3)−P(X=4)=1−15C0(0.8)0(0.2)15−15C1(08)1

(0.2)14.−15C4(0.8)4(0.2)4=10

All of them are extroverts

P(X=15)=15C15(0.8)15(0.2)0=15!0!(15)∗(0.8)15∗(0.2)0=0.035

Answer 2

Step-by-step explanation:

Answers

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Step-by-step explanation:

Let 'X' be a random variable representing the number of extroverts.

Given approximately 80% of all marketing personnel are extroverts, hence, probability of success will be

p=0.80andq=1−0.80=0.2.

Given, in a meeting there are 15 marketing personnel present.

Then

X∼Bin(15,0.8)

(a) at least 10 are extroverts.

P(X≥10)=P(X=10)+P(X=11)+P(X=12)+P(X=13)P(X=14)+P(X=15)=15C10(0.8)10(0.2)5+15C15(0.8)15(0.2)0=0.939

At least 5 are extroverts.

P(X≥5)=1−P(X=0)−P(X=1)−P(X=2)−P(X=3)−P(X=4)=1−15C0(0.8)0(0.2)15−15C1(08)1

(0.2)14.−15C4(0.8)4(0.2)4=10

All of them are extroverts

P(X=15)=15C15(0.8)15(0.2)0=15!0!(15)∗(0.8)15∗(0.2)0=0.035