Question

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 1

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,

$P(x)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

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 = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

$P(X$

$=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}$

$=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}$

$=0.1971110499$

$\approx 0.1971$