Mathematics, 17.03.2020 03:35 ansarishaheer2888

A company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall. a. What is the probability that the inspection procedure will pass the shipment? b. What is the expected number of defectives in this process of inspecting 5 items? c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?

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Answer from: zekrader18

(a) The probability that the inspection procedure will pass the shipment is 0.9185.

(b) The expected number of defectives in this process of inspecting 5 items is 0.50.

Step-by-step explanation:

Let X = number of defective items.

The probability of selecting a defective item is, p = 0.10.

A sample of n = 5 items is selected at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.10.

The probability mass function of X is:

$P(X=x)={5\choose x}0.10^{x}(1-0.10)^{5-x};\ x=0,1,2,3....$

It is provided that the shipment will pass if there are no more than 1 defective items in the selected 5 units.

(a)

Compute the probability that the inspection procedure will pass the shipment as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

$={5\choose 0}0.10^{0}(1-0.10)^{5-0}+{5\choose 1}0.10^{1}(1-0.10)^{5-1}\\=(1\times 1\times 0.59049) + (5\times 0.10\times 0.6561)\\=0.59049+0.32805\\=0.91854\\\approx0.9185$

Thus, the probability that the inspection procedure will pass the shipment is 0.9185.

(b)

The expected value of a Binomial distribution is:

E(X)=np

Compute the expected number of defectives in this process of inspecting 5 items as follows:

$E(X)=5\times0.10=0.50$

Thus, the expected number of defectives in this process of inspecting 5 items is 0.50.

(c)

The probability of finding a defective is now changed to 0.20.

Compute the probability that there will be 4 defectives in a sample of 5 as follows;

$P(X=4)={5\choose 4}0.20^{4}(1-0.20)^{5-4}\\=5\times0.0016\times0.20\\=0.0016$

Thus, the probability that there will be 4 defectives in a sample of 5 is 0.0016.

Answer from: Quest

Best answer that i found on yahoo ok set : x= length of 2nd piece in cm x- 3= length of 1st piece in cm x-4= length of 3rd piece in cm then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20 3x - 7= 20 3x= 27 x= 9 the shortest piece is the third piece which is x- 4 x-4 9-4 = 5 the length of the smallest piece is 5 cm. the other lengths are 9 cm 6 cm

Answer from: Quest

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points

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