Psychology: Myers Briggs Approximately 75% of all marketing personnel are extrov

Question

Psychology: Myers Briggs Approximately 75% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are intro- verts (Source: A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley). At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? In a group of 5 computer programmers, what is the probability that none are introverts? What is the probability that 3 or more are introverts? What is the probability that all are introverts?

Explanation / Answer

Solution:-

Binomial distribution is to be used
P(x) = nCx * px * q(n-x)
(a)
p= 0.75, q = 1 - p = 0.25
n = 15
x = 10 or more

P(x>=10)= P(x=10) + P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)
= 15C10 * (0.75)10 * (0.25)5 + 15C11 * (0.75)11 * (0.25)4 + 15C12 * (0.75)12 * (0.25)3 + 15C13 * (0.75)13 * (0.25)2
+ 15C14 * (0.75)14 * (0.25)1 + 15C15 * (0.75)15 * (0.25)0
= 0.8516(Ans.)

ii) Required probability = P(x>/= 5)
= 1 - [P(x=0)+P(x=1)+... +P(x=4)]

= 1 - [15C0 * (0.75)0 * (0.25)15 + 15C1 * (0.75)1 * (0.25)14 +  15C2 * (0.75)2 * (0.25)13 +  15C3 * (0.75)3 * (0.25)12 +  15C4 * (0.75)4 * (0.25)11]

= 1 - [1 * (0.25)15 + 15 * (0.75)1 * (0.25)14 + 105 * (0.75)2 * (0.25)13 + 455 * (0.75)3 * (0.25)12 + 1365 * (0.75)4 * (0.25)11 ]

iii) Required probability = P(x=15)
= 15C15*0.7515*0.250
= 0.01336

b).

p = 0.60

q = 1 - p = 0.40

n= 5

i) Required probability = P(x=0) = 5C0*0.60*0.45

= 1 . 1 . 0.01024 = 0.01024

ii) Required probability = P(x >/= 3)

= P(x=3)+P(x=4)+P(x=5)

=  5C3*0.63*0.42 + 5C4*0.64*0.41 +  5C5*0.65*0.40

= 10 * 0.03456 + 5 * 0. 0.05184 + 1 * 0.07776

iii) Required probability = P(x=5) = 5C5*0.65*0.40

= 1.0.07776.1 = 0.07776

Kindly work out the calculations left using a scientific calculator.

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