In a shipment of 2,000 beach balls 150 are defective the manufacturer generates a random sample to simulate 20 beach balls to inspect in the next shipment the integers 1 to 150 represent defective beach balls

The concluding part of the question as obtained from the textbook;

The number on the 20 beach balls that come up in the simulated sample:

42, 1701, 638, 397, 113, 1243, 912, 380, 769, 1312, 76, 547, 721, 56, 4, 1411, 1766, 677, 201, 1840

A) Based on this sample, how many

defective beach balls might the

manufacturer expect in the next

shipment?

B) What is the difference between the

number of defective beach balls in the

actual shipment and the number

predicted in the next shipment?

A) 500 defective beach balls.

B) Difference between the

number of defective beach balls in the

actual shipment and the number

predicted in the next shipment = 350

Step-by-step explanation:

The beach balls are labelled 1 to 2000 with the 150 defective ones labelled 1 to 150.

Then a random sample of 20 beach balls is picked, and the numbers are presented as

42, 1701, 638, 397, 113, 1243, 912, 380, 769, 1312, 76, 547, 721, 56, 4, 1411, 1766, 677, 201, 1840

Note that only the defective beach balls have numbers 1 to 150.

A) The number of beach balls with numbers from 1 to 150 in the sample is 5 (numbers 42, 113, 76, 56, 4). This is the number of defective beach balls in the sample.

Probability of getting a defective ball in the next shipment = (5/20) = 0.25

If every shipment contains 2000 beach balls, then there will be (0.25 × 2000) defective beach balls in the next sample; 500 defective beach balls.

B) Number of defective beach balls in actual shipmemt = 150

Number of predicted defective beach balls in the next shipment = 500

difference = 500 - 150 = 350.

Hope this Helps!!!

yes

step-by-step explanation:

they have to be opposite and reciprocal of eachother.

      2

  2 is to go somewhat to the top right of a because it is squared    

v= a h/3