Summarizing numerical data sets - HiSET

First, we test to see if she is already assured of a 70 average. The worst-case scenario for Janice is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Janice is already assured of a mean of at least 70, since any score can only raise her grade.

First, we test to see if she is already assured of a 80 average. The worst-case scenario for Donna is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Donna has not yet attained her 80. Now, we test to determine what her twelfth test score must be. If she scores 60 or less she will drop this score as well as the 54, so we will assume that she scores more than 60. Calling this score , the mean of her scores will be the the sum of her best nine scores thus far and this unknown score , divided by 10. The mean should be greater than or equal to 80, so we can set up and solve for in this inequality:

Note that the third (60) and fifth (54) scores have been omitted. Add the known scores to get

Multiply both sides by 10:

Subtract 656 from both sides:

Donna would have to score 144 or more on the twelfth test - an impossible feat. She cannot attain an average of 80 or greater.

First, we test to see if she is already assured of a 70 average. The worst-case scenario for Janice is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Janice is already assured of a mean of at least 70, since any score can only raise her grade.

First, we test to see if she is already assured of a 80 average. The worst-case scenario for Donna is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Donna has not yet attained her 80. Now, we test to determine what her twelfth test score must be. If she scores 60 or less she will drop this score as well as the 54, so we will assume that she scores more than 60. Calling this score , the mean of her scores will be the the sum of her best nine scores thus far and this unknown score , divided by 10. The mean should be greater than or equal to 80, so we can set up and solve for in this inequality:

Note that the third (60) and fifth (54) scores have been omitted. Add the known scores to get

Multiply both sides by 10:

Subtract 656 from both sides:

Donna would have to score 144 or more on the twelfth test - an impossible feat. She cannot attain an average of 80 or greater.

First, we test to see if she is already assured of a 70 average. The worst-case scenario for Janice is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Janice is already assured of a mean of at least 70, since any score can only raise her grade.

First, we test to see if she is already assured of a 80 average. The worst-case scenario for Donna is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Donna has not yet attained her 80. Now, we test to determine what her twelfth test score must be. If she scores 60 or less she will drop this score as well as the 54, so we will assume that she scores more than 60. Calling this score , the mean of her scores will be the the sum of her best nine scores thus far and this unknown score , divided by 10. The mean should be greater than or equal to 80, so we can set up and solve for in this inequality:

Note that the third (60) and fifth (54) scores have been omitted. Add the known scores to get

Multiply both sides by 10:

Subtract 656 from both sides:

Donna would have to score 144 or more on the twelfth test - an impossible feat. She cannot attain an average of 80 or greater.

First, we test to see if she is already assured of a 70 average. The worst-case scenario for Janice is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Janice is already assured of a mean of at least 70, since any score can only raise her grade.

First, we test to see if she is already assured of a 80 average. The worst-case scenario for Donna is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

Donna has not yet attained her 80. Now, we test to determine what her twelfth test score must be. If she scores 60 or less she will drop this score as well as the 54, so we will assume that she scores more than 60. Calling this score , the mean of her scores will be the the sum of her best nine scores thus far and this unknown score , divided by 10. The mean should be greater than or equal to 80, so we can set up and solve for in this inequality:

Note that the third (60) and fifth (54) scores have been omitted. Add the known scores to get

Multiply both sides by 10:

Subtract 656 from both sides:

Donna would have to score 144 or more on the twelfth test - an impossible feat. She cannot attain an average of 80 or greater.

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

Simplify the numerator by combining the like terms:

Now, split the fraction, and reduce to lowest terms:

,

the correct response.

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

Only in one case does it hold that is a prime integer.

The correct response is one - .

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

Simplify the numerator by combining the like terms:

Now, split the fraction, and reduce to lowest terms:

,

the correct response.

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

Only in one case does it hold that is a prime integer.

The correct response is one - .

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

Simplify the numerator by combining the like terms:

Now, split the fraction, and reduce to lowest terms:

,

the correct response.

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

Only in one case does it hold that is a prime integer.

The correct response is one - .

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 29 already occurs three times in the data set. For the set to be bimodal, must be equal to one of the values that occurs once - 17, 21, 27, 35, or 37. Since it is given that is prime - having only two factors, 1 and itself - can only be either of 17 and 37, the other three values having other factors.

The correct response is two.

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

Only in one case does it hold that is a prime integer.

The correct response is one - .

The median of a set of nine data values - an odd number - is the value that appears in the middle when the values are ranked. For 8 to be the median, 8 must be in the middle - that is, four values must appear before 8, and four values must appear after 8.

In the given data set, four values - 4, 5, 5, 6 - are known to be appear before 8, so must appear after 8. Since is an integer from 1 to 10, can only be 8, 9, or 10. This makes three the correct response.

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

Simplify the numerator by combining the like terms:

Now, split the fraction, and reduce to lowest terms:

,

the correct response.

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 29 already occurs three times in the data set. For the set to be bimodal, must be equal to one of the values that occurs once - 17, 21, 27, 35, or 37. Since it is given that is prime - having only two factors, 1 and itself - can only be either of 17 and 37, the other three values having other factors.

The correct response is two.

The median of a set of nine data values - an odd number - is the value that appears in the middle when the values are ranked. For 8 to be the median, 8 must be in the middle - that is, four values must appear before 8, and four values must appear after 8.

In the given data set, four values - 4, 5, 5, 6 - are known to be appear before 8, so must appear after 8. Since is an integer from 1 to 10, can only be 8, 9, or 10. This makes three the correct response.