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Suppose that, on average, 4 percent of all CD drives received by a computer company are defective.
The company has adopted the following policy:
Sample 50 CD drives in each shipment, and accept the shipment if none are defective.
Using this information, determine the following:
a) What fraction of shipments will be accepted?
b) If the policy changes so that a shipment is accepted if only one CD drive in the sample isdefective, what fraction of shipments will be accepted?
c) What is the probability that a sample size of 50 will contain at least 10 defective CD drives?

Hint: You can use BINOMDIST function.

Please help me, this is an excel question

Respuesta :

gauravsingh23VT

The percentage of packages that will be accepted is 0.1299. as, The percentage of packages that will be accepted is 0.1299.

What is the method of doing probability?

Probability of a defective DVD = 0.04

Using Binomial distribution,

Fraction of shipments accepted = Probability of zero defects in sample of 50 = P(X = 0)

= 50C0 * 0.040 * (1 - 0.04)⁵⁰

= 0.96⁵⁰

= 0.1299

Fraction of shipments accepted = Probability of zero or one defects in sample of 50 = P(X = 0) + P(X = 1)

= 50C0 * 0.040 * (1 - 0.04)50-0 + 50C1 * 0.041 * (1 - 0.04)50-1

= 0.9650 + 50 * 0.04 * 0.9649

= 0.4005

Thus, it is 0.1299.

For more details about method of doing probability, click here:

https://brainly.com/question/12868212

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