In a certain process, the probability of producing an defective component is 0.07.

a. In a sample of 250 randomly chosen components, what is the probability that fewer than 20 of them are defective?

b. In a sample of 10 randomly chosen components, what is the probability that one or more of them is defective?

c. To what value must the probability of an defective component be reduced so that only 1% of lots of 250 components contain 20 or more that are defective?

Answer:

Step1 of 6:

Given, the probability of producing a defective component is 0.07.

a). The aim is to find the probability that fewer than 20 of them are defective.

Let X be the number of components in a sample of 250 that are defective.

Then,

Where, n = 250, p = 0.07.

Now, compute mean

= (250)(0.07)

= 17.5

And standard deviation (where, q = 1- p = 1- 0.07 = 0.93)

=

= 4.0342

Step 2 of 6:

Now, to find Here we using continuity correction factor.

Therefore,

= P(X< 20 - 0.5)

= P(X < 19.5)

Now, the z-score is

= 0.50)

= 0.6915. ( this value from statistical table)

Conclusion:

Therefore, the probability that fewer than 20 of them are defective is 0.6915.

Step 3 of 6:

b). The aim is to find the probability that one or more of them is defective.

Let y follows the number of components in a sample of 10 that are defective.

That is,

Where, n = 10 , p = 0.07.

Now, to find

= (q = 1 - p)

= 1- 0.4839

= 0.5160

Conclusion:

Therefore, the probability that one or more of them defective is 0.5160.