It has been estimated that about 30% of frozen chicken contain enough salmonella bacteria to cause illness if improperly cooked. A consumer purchases 12 frozen chickens. What is the probability that the consumer will have more than 6 contaminated chickens? a) .961.
b) .118.
c) .882.
d) .039.
Refer to the previous question. Suppose that a supermarket buys 1000 frozen chickens from a supplier. The number of frozen chickens that may be contaminated that arewithin two standard deviations of the mean is between A and B. The numbers A and B are:.
a) (90,510).
b) (290.8,309.2).
c) (0,730).
d) (271,329).
e) (255,345).

0.118 ; (271, 329)

Step-by-step explanation:

Given;

Probability of success , p = 0.3

Number of trials, n = 12

P(x ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)

Using binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Using calculator :

P(x ≥ 6) = 0.1178

= 0.118

B.)

Suppose number of trials, n = 1000

And number of defective chickens is within 1 standard deviation of the mean :

Mean = np ; p = 0.3

Mean = 1000 * 0.3 = 300

Standard deviation (s) : sqrt(np(1-p))

s = sqrt(1000 * 0.3 * 0.7)

s = sqrt(210)

s = 14.49

2 standard deviations of the mean:

Mean ± 2s

Lower bound : 300 - 2(14.49) = 271.02

Upper bound : 300 + 2(14.49) = 328.98

(271.02, 328.98) = (271, 329)

10 you would take 50/ .2

step-by-step explanation: