The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more than 20 pounds. A sample of 16 babies is studied. What is the probability that fewer than 3 weigh more than 20 pounds?

Answer:0.1971 ( approx )Step-by-step explanation:Let X represents the event of weighing more than 20 pounds,Since, the binomial distribution formula is,[tex]P(x)=^nC_r p^r q^{n-r}[/tex]Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]Given,The probability of weighing more than 20 pounds, p = 25% = 0.25,⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75Total number of samples, n = 16,Hence, the probability that fewer than 3 weigh more than 20 pounds,[tex]P(X<3) = P(X=0)+P(X=1)+P(X=2)[/tex][tex]=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}[/tex][tex]=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}[/tex][tex]=0.1971110499[/tex][tex]\approx 0.1971[/tex]