The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of twelve syringes taken from the batch. Suppose the batch contains 5% defective syringes.

(b) Find ?. (Enter your answer to two decimal places.)
What is the expected number of defective syringes the inspector will find? (Enter your answer to two decimal places.)
(c) What is the probability that the batch will be accepted? (Round your answer to three decimal places.)
(d) Find ?. (Round your answer to three decimal places.)

Step-by-step explanation:

p = 0.05 ( or 5%)

n = 12

Part b)

Expected value = n X p

expected value = 12 X 0.05

Expected value = 0.60

Part c)

Batch will be accepted if the number of defectives is less than 2

P(accepted) = P( x = 0) + P (x = 1)

where x = number of defectives

From binomial formula;

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

we got n = 12, p = 0.05

P(x=0) = 12C0 X 0.05^0 X (0.95)^12

P(x=0) = 0.54036

P(x=1) = 12C1 X (0.05)61 X (0.95)^11

P(x=1) = 0.34128

P(accepted) = 0.54036 + 0.34128

P(accepted) = 0.88164

Part d)

Standard deviation = √(0.05 X 0.95/12) = 0.062915

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.