forty - five percent of consumers say it is important that the clothing they buy is made without child…

forty - five percent of consumers say it is important that the clothing they buy is made without child labor. you randomly select 16 consumers. find the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is (a) exactly three, (b) at least ten, and (c) less than eleven.\n(a) the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is exactly three is p(exactly three)=0.022. (round to three decimal places as needed.)\n(b) the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is at least ten is p(at least ten)=. (round to three decimal places as needed.)

forty - five percent of consumers say it is important that the clothing they buy is made without child labor. you randomly select 16 consumers. find the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is (a) exactly three, (b) at least ten, and (c) less than eleven.\n(a) the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is exactly three is p(exactly three)=0.022. (round to three decimal places as needed.)\n(b) the probability that the number of consumers who say it is important that the clothing they buy is made without child labor is at least ten is p(at least ten)=. (round to three decimal places as needed.)

Answer

Answer:

(a) 0.022 (b) 0.102 (c) 0.898

Explanation:

Step1: Identify the binomial formula

$P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 16$ (number of trials), $p=0.45$ (probability of success), $C(n,k)=\frac{n!}{k!(n - k)!}$

Step2: Calculate (a)

For $k = 3$: $C(16,3)=\frac{16!}{3!(16 - 3)!}=\frac{16\times15\times14}{3\times2\times1}=560$ $P(X = 3)=560\times(0.45)^{3}\times(1 - 0.45)^{16 - 3}$ $P(X = 3)=560\times0.45^{3}\times0.55^{13}\approx0.022$

Step3: Calculate (b)

$P(X\geq10)=\sum_{k = 10}^{16}C(16,k)\times(0.45)^{k}\times(0.55)^{16 - k}$ $C(16,10)=\frac{16!}{10!(16 - 10)!}=8008$ $C(16,11)=\frac{16!}{11!(16 - 11)!}=4368$ $C(16,12)=\frac{16!}{12!(16 - 12)!}=1820$ $C(16,13)=\frac{16!}{13!(16 - 13)!}=560$ $C(16,14)=\frac{16!}{14!(16 - 14)!}=120$ $C(16,15)=\frac{16!}{15!(16 - 15)!}=16$ $C(16,16)=\frac{16!}{16!(16 - 16)!}=1$ $P(X\geq10)=8008\times(0.45)^{10}\times(0.55)^{6}+4368\times(0.45)^{11}\times(0.55)^{5}+1820\times(0.45)^{12}\times(0.55)^{4}+560\times(0.45)^{13}\times(0.55)^{3}+120\times(0.45)^{14}\times(0.55)^{2}+16\times(0.45)^{15}\times(0.55)^{1}+1\times(0.45)^{16}\times(0.55)^{0}\approx0.102$

Step4: Calculate (c)

$P(X\lt11)=1 - P(X\geq11)$ $P(X\geq11)=\sum_{k = 11}^{16}C(16,k)\times(0.45)^{k}\times(0.55)^{16 - k}$ $P(X\lt11)=1-(P(X = 11)+P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15)+P(X = 16))$ $P(X\lt11)\approx1 - 0.102=0.898$