Question

At a grocery store the manager randomly selects 20 cartons of eggs, he finds 3 cartons with at least 1 broken egg. If the refrigerator holds 500 cartons, how many would you expect to have at least 1 broken egg?
A shipment to a warehouse consists of 3700 DVD players. The manager chooses random sample of 50 DVD players and finds 3 are defective. How many DVD players are likely to be defective?

Answer 1

Answer:

You would expect 75 cartons with at least 1 broken egg.

There are likely to be 222 DVD players that are defective.

Step-by-step explanation:

Both examples given in this question represent a ratio of broken or defective to a sample number of cartons or DVD players.  In order to find how many would be broken or defective overall, you can set up equivalent ratios, or proportions.

Egg Cartons: $\frac{broken}{#cartons}=\frac{3}{20}=\frac{x}{500}$

'x' represents the number of cartons out of 500 that would be expected to have at least 1 broken egg.  To solve for 'x' you can cross-multiply and divide, or you can find what factor you would multiply the denominator by to get 500 and then multiply the same factor by the numerator to find 'x'.  In this case 500/20 = 25, so the factor is 25.  3 x 25 = 75, so you have 75 expected cartons with a broken egg.

DVD Players:  $\frac{defective}{#players}=\frac{3}{50}=\frac{x}{3700}$

'x' represents the number of DVD players out of 3700 that would be expected to be defective.  To solve for 'x' you can cross-multiply and divide, or you can find what factor you would multiply the denominator by to get 3700 and then multiply the same factor by the numerator to find 'x'.  In this case 3700/50 = 74, so the factor is 74.  3 x 74 = 222, so you have 22 likely defective DVD players.