Respuesta :
Answer:
The probability that 10 or more are extroverts is
Step-by-step 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;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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 ~ [tex]Binom(n=15, p=0.80)[/tex]
Now, Probability that 10 or more are extroverts is given by = P(X [tex]\geq[/tex] 10)
P(X [tex]\geq[/tex] 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
= [tex]\binom{15}{10}\times 0.80^{10}\times (1-0.80)^{15-10}+ \binom{15}{11}\times 0.80^{11}\times (1-0.80)^{15-11}+ \binom{15}{12}\times 0.80^{12}\times (1-0.80)^{15-12}+ \binom{15}{13}\times 0.80^{13}\times (1-0.80)^{15-13}+ \binom{15}{14}\times 0.80^{14}\times (1-0.80)^{15-14}+ \binom{15}{15}\times 0.80^{15}\times (1-0.80)^{15-15}[/tex]
= [tex]3003 \times 0.80^{10} \times 0.20^{5}+ 1365 \times 0.80^{11} \times 0.20^{4}+ 455 \times 0.80^{12} \times 0.20^{3}+ 105 \times 0.80^{13} \times 0.20^{2}+ 15 \times 0.80^{14} \times 0.20^{1}+ 1 \times 0.80^{15} \times 0.20^{0}[/tex]
= 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.
