Question

a public health organization reports that 40% of baby boys 6-8 months old in the US weigh more than 20 lbs.a sample of 10 babies is studied.
1. what is probability of exactly 4 of them weigh more tab 20 lbs
2. what is probability that more than 4 weigh more than 20 lbs
3. what is probability that fewer than 3 weign more than 20 lbs

Answer 1

Answer:

Given,

sample n = 10

p = 0.40

q = 1 - p

= 1 - 0.40

= 0.60

Consider,

Binomial distribution P(X) = nCr*p^r*q^(n-r)

nCr = n!/(n-r)!*r!

a)

P(X = 4) = 10C4*0.40^4*0.60^6

= 210*0.4^4*0.6^6

= 0.2508

b)

P(X > 4) = 1 - P(X <= 4)

= 1 - [P(0) + P(1) + P(2) + P(3) + P(4)]

= 1 - [10C0*0.4^0*0.6^10 + 10C1*0.4^1*0.6^9 + 10C2*0.4^2*0.6^8 + 10C3*0.4^3*0.6^7 + 10C4*0.4^4*0.6^6]

= 1 - [0.0060 + 0.0403 + 0.1209 + 0.2150 + 0.2508]

= 1 - 0.6330

= 0.3670

c)

P(X < 3) = P(0) + P(1) + P(2)

= 10C0*0.4^0*0.6^10 + 10C1*0.4^1*0.6^9 + 10C2*0.4^2*0.6^8

= 0.0060 + 0.0403 + 0.1209

= 0.1672