Problem 1 :
A shipment to a warehouse consists of 3,500 MP3 players. The manager chooses a random sample of 50 MP3 players and finds that 3 are defective. How many MP3 players in the shipment are likely to be defective?
Problem 2 :
Based on the information in question 1, how many MP3 players in the shipment can we predict to be damaged if 6 MP3s in the sample had been damaged?
Problem 3 :
How could we use estimation to check our answer for question 2 is reasonable?
Problem 4 :
A manager samples the receipts of every fifth person who goes through the line. Out of 50 people, 4 had a mispriced item. If 600 people go to this store each day, how many people can we expect to have a mispriced item?
1. Answer :
It is reasonable to make a prediction about the population, because this sample is random.
Step 1 :
Set up a proportion.
Step 2 :
Substitute values into the proportion.
Substitute known values. Let x be the number of defective MP3 players in the population.
3/50 = x/3500
Least common multiple of (50, 70) = 3500.
So multiply the numerator and denominator of the fraction 3/50 by 70.
(3 ⋅ 70)/(3 ⋅ 70) = x/3500
210/3500 = x/3500
210 = x
Based on the sample, you can predict that 210 MP3 players in the shipment would be defective.
2. Answer :
Step 1 :
Set up a proportion.
Step 2 :
Substitute values into the proportion.
Substitute known values. Let x be the number of defective MP3 players in the population.
6/50 = x/3500
(6 ⋅ 70)/(50 ⋅ 70) = x/3500
420/3500 = x/3500
420 = x
Based on the sample, we can predict that 420 MP3 players in the shipment would be defective.
3. Answer :
6 is a little more than 10% of 50.
10% of 3,500 is 350, and 420 is a little more than that.
4. Answer :
Step 1 :
Set up a proportion.
Step 2 :
Substitute values into the proportion.
Substitute known values. Let x be the number of mispriced items in the population.
4/50 = x/600
Least common multiple of (50, 600) = 600.
So, multiply the numerator and denominator of the fraction 4/50 by 12.
(4 ⋅ 12)/(50 ⋅ 12) = x/600
48/600 = x/600
48 = x
Based on the sample, we can expect 48 mispriced items in the population of 600 people.
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