(7+5) ÷ 6 + 10 to the second power
1 answer:
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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?
Answer:
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.
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