Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. find the probability of a household having: a. 1 or 9 radios b. 7 or fewer radios c. 5 or more radios d. fewer than 9 radios e. more than 7 radios

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x=r) = nCr × q^(n-r) × p^r

From the information given,

n = 11

p = 40% = 40/100 = 0.4

q = 1 - 0.4 = 0.6

x represent the number of radios

a) P( x = 1) or P(x = 9)

P(x = 1) = 11C1 × 0.6^(11-1) × 0.4^1

P(x = 1) = 0.027

P(x = 9) = 11C9 × 0.6^(11-9) × 0.4^9

P(x = 9) = 0.0052

P( x = 1) or P(x = 9) = 0.027 + 0.0052 = 0.0322

b) P(x lesser than or equal to 7) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)

P(x = 0) = 11C0 × 0.6^(11-0) × 0.4^0 = 0.004

P(x = 1) = 0.027

P(x = 2) = 11C2 × 0.6^(11-2) × 0.4^2 = 0.089

P(x = 3) = 11C3 × 0.6^(11-3) × 0.4^3 = 0.177

P(x = 4) = 11C4 × 0.6^(11-4) × 0.4^4 = 0.24

P(x = 5) = 11C5 × 0.6^(11-5) × 0.4^5 = 0.22

P(x = 6) = 11C6 × 0.6^(11-6) × 0.4^6 = 0.15

P(x = 7) = 11C7 × 0.6^(11-7) × 0.4^7 = 0.15

P(x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

c) P(x greater than or equal to 5) = 1 - P(x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

d) P(x lesser than 9) = P(x lesser than or equal to 7) + P(x = 8)

P(x = 8) = 11C8 × 0.6^(11-8) × 0.4^8 = 0.02

P(x lesser than 9) = 0.977 + 0.02 = 0.997

e. P(x greater than 7) = 1 - P(x lesser than or equal to 7) = 1 - 0.977 = 0.023

use distributive property on all like terms and change your signs if you have a double negative or if you have positive/negative