In a study, 37% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to nuclear plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health.

The probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.2946

Step-by-step explanation:

BIONOMIAL DISTRIBUTION

pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k

where

k = number of successes in trials    

n = is the number of independent trials    

p = probability of success on each trial

I.

mean = np

where

n = total number of repetitions experiment is excueted

p = success probability

mean = 12 * 0.37

= 4.44

II.

variance = npq

where

n = total number of repetitions experiment is excueted

p = success probability

q = failure probability

variance = 12 * 0.37 * 0.63

= 2.7972

III.

standard deviation = sqrt( variance ) = sqrt(2.7972)

=1.6725

the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health

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

= ( 12 3 ) * 0.37^3 * ( 1- 0.37 ) ^9 + ( 12 2 ) * 0.37^2 * ( 1- 0.37 ) ^10 + ( 12 1 ) * 0.37^1 * ( 1- 0.37 ) ^11 + ( 12 0 ) * 0.37^0 * ( 1- 0.37 ) ^12    

= 0.2947

google dat sis

step-by-step explanation:

other answer is correct. i had used the radius of the radius somehow. a website: