Question

Thirty-nine percent of U.S. Adults think that the government should help fight childhood obesity. You randomly select six U.S. Adults. Find the probability that the number of U.S. Adults who think that the

Answer 1

we are to find: the probability that the number of U.S. adult who thinks that the government should help fight childhood obesity is:

(a) exactly two (b) at least four

Answer:

P(X =2) = 0.3160

P(X ≥ 4) = 0.1656

Explanation:

From the given information:

The population proportion = 39/100 = 0.39

the sample size = 6

(a) The required probability that exactly two adults fight childhood obesity out of the six is:

`P(X=2) = (^6_2) (0.39)^2 (1-0.39)^(6-2)`

`P(X=2) = (6!)/(2!(6-2)!) (0.39)^2 (1-0.39)^(4)`

`P(X=2) = 15 * 0.1521 * 0.1385`

P(X =2) = 0.3160

(b)

The probability that at least 4 adults did is:

P( X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

`P( X \ge 4) =(^6_4) (0.39)^4 (1-0.39)^(6-4) + (^6_5) (0.39)^5 (1-0.39)^(6-5)+ (^6_6) (0.39)^6 (1-0.39)^(6-6)`

`P( X \ge 4) =((6!)/(4!(6-4)!)) (0.39)^4 (1-0.39)^(2) +((6!)/(5!(6-5)!)) (0.39)^5 (1-0.39)^(1)+ ((6!)/(6!(6-6)!)) (0.39)^6 (1-0.39)^(0)`

P(X ≥ 4) = 0.1291 + 0.0330 + 0.0035

P(X ≥ 4) = 0.1656