Question

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 55% 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

The probability that 10 or more are extroverts is

Explanation:

We are given that approximately 80% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts.

Also, a sample of 15 marketing personnel is chosen.

The above situation can be represented through Binomial distribution;

`P(X=r) = \binom{n}{r}p^(r) (1-p)^(n-r) ; x = 0,1,2,3,.....`

where, n = number of trials (samples) taken = 15 marketing personnel

r = number of success = 10 or more

p = probability of success which in our question is % of marketing

personnel that are extroverts, i.e; 80%

LET X = Number of marketing personnel that are extroverts

So, it means X ~
`Binom(n=15, p=0.80)`

Now, Probability that 10 or more are extroverts is given by = P(X
`\geq` 10)

P(X
`\geq` 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

=
`\binom{15}{10}* 0.80^(10)* (1-0.80)^(15-10)+ \binom{15}{11}* 0.80^(11)* (1-0.80)^(15-11)+ \binom{15}{12}* 0.80^(12)* (1-0.80)^(15-12)+ \binom{15}{13}* 0.80^(13)* (1-0.80)^(15-13)+ \binom{15}{14}* 0.80^(14)* (1-0.80)^(15-14)+ \binom{15}{15}* 0.80^(15)* (1-0.80)^(15-15)`

=
`3003 * 0.80^(10) * 0.20^(5)+ 1365 * 0.80^(11) * 0.20^(4)+ 455 * 0.80^(12) * 0.20^(3)+ 105 * 0.80^(13) * 0.20^(2)+ 15 * 0.80^(14) * 0.20^(1)+ 1 * 0.80^(15) * 0.20^(0)`

= 0.1032 + 0.1876 + 0.2501 + 0.2309 + 0.1319 + 0.0352 = 0.9389

So, the probability that 10 or more are extroverts is 0.9389.