Math 1 Summary Statistics Quiz

1. Erin and Jessica both took the same four health tests. Their grades are listed below. Which of the following statements is true? Erin: 89, 76, 92, 83 Jessica: 95, 69, 78, 87

Erin's mean score is lower than Jessica's mean score.

Jessica's mean score is lower than Erin's mean score.

Jessica's median score is higher than Erin's median score.

Jessica and Erin had the same median score.

To find the mean score, we add up all the scores and divide by the number of tests taken. Erin's mean score is (89 + 76 + 92 + 83) / 4 = 85. Jill's mean score is (95 + 69 + 78 + 87) / 4 = 82. Therefore, Erin's mean score is higher than Jessica's mean score.

Explanation

To find the mean score, we add up all the scores and divide by the number of tests taken. Erin's mean score is (89 + 76 + 92 + 83) / 4 = 85. Jill's mean score is (95 + 69 + 78 + 87) / 4 = 82. Therefore, Erin's mean score is higher than Jessica's mean score.

2. A box-and-whisker diagram is the most appropriate diagram to use when trying to determine the __________ of a data set.

Mean

Spread

Mode

Mean Absolute Deviation

A box-and-whisker diagram is the most appropriate diagram to use when trying to determine the spread of a data set. This is because the diagram displays the minimum, maximum, median, and quartiles of the data, providing a visual representation of how the data is distributed and the range between the lowest and highest values. By examining the length of the box and the length of the whiskers, one can easily determine the spread or variability of the data.

Explanation

A box-and-whisker diagram is the most appropriate diagram to use when trying to determine the spread of a data set. This is because the diagram displays the minimum, maximum, median, and quartiles of the data, providing a visual representation of how the data is distributed and the range between the lowest and highest values. By examining the length of the box and the length of the whiskers, one can easily determine the spread or variability of the data.

3. Mike is deciding on tile for his bathroom. He has narrowed it down to 7 tiles. The price of the 7 tiles are listed below. Find the mean price of the tiles. $3.29, $4.15, $2.97, $2.61, $5.00, $2.99, $3.75

$24.76

$3.54

$3.29

$3.53

The mean price of the tiles can be found by adding up all the prices of the tiles and then dividing by the total number of tiles. In this case, the sum of the prices is $24.76 and there are 7 tiles. Dividing $24.76 by 7 gives us $3.54, which is the mean price of the tiles.

Explanation

The mean price of the tiles can be found by adding up all the prices of the tiles and then dividing by the total number of tiles. In this case, the sum of the prices is $24.76 and there are 7 tiles. Dividing $24.76 by 7 gives us $3.54, which is the mean price of the tiles.

4. Justin took 10 math tests. His scores are listed below. Which of the following is the correct description of the lower 25% of Justin's scores? 82, 89, 75, 91, 83, 82, 71, 65, 93, 83

89 - 93

82.5 - 93

65 - 82.5

65 - 75

The correct description of the lower 25% of Justin's scores is 65 - 75. This means that 25% of his scores fall between 65 and 75.

Explanation

The correct description of the lower 25% of Justin's scores is 65 - 75. This means that 25% of his scores fall between 65 and 75.

5. Find the interquartile range for the data set below. S = {1220, 1890, 1560, 1670, 1180, 2100, 1440}

280

370

1560

1580

The interquartile range is a measure of the spread of the middle 50% of the data. To find it, we first need to find the first quartile (Q1) and the third quartile (Q3). To do this, we arrange the data set in ascending order: 1180, 1220, 1440, 1560, 1670, 1890, 2100. Q1 is the median of the lower half of the data set, which is 1500. Q3 is the median of the upper half of the data set, which is 1780. The interquartile range is then calculated by subtracting Q1 from Q3: IQR = Q3 - Q1 = 1780 - 1500 = 280.

Explanation

The interquartile range is a measure of the spread of the middle 50% of the data. To find it, we first need to find the first quartile (Q1) and the third quartile (Q3). To do this, we arrange the data set in ascending order: 1180, 1220, 1440, 1560, 1670, 1890, 2100. Q1 is the median of the lower half of the data set, which is 1500. Q3 is the median of the upper half of the data set, which is 1780. The interquartile range is then calculated by subtracting Q1 from Q3: IQR = Q3 - Q1 = 1780 - 1500 = 280.

6. The heights of 12 tomato plants in centimeters are listed below. In general the average height of a tomato plant is 90 cm and the median height is 86.5 cm. How does this data set compare? 67, 76, 79, 84, 86, 88, 89, 90, 91, 95, 95, 100

The mean for this data set is lower and the median is higher.

The mean for this data set is higher and the median is lower.

Both the mean and the median for this data set is lower.

Both the mean and median for this data set is higher.

The given data set of tomato plant heights has a lower mean and a higher median compared to the general average height of tomato plants. This suggests that there are a few plants with significantly higher heights in the data set, which pulls up the median, but the majority of plants have heights closer to the lower end, bringing down the mean. Therefore, the mean is lower and the median is higher in this data set.

Explanation

The given data set of tomato plant heights has a lower mean and a higher median compared to the general average height of tomato plants. This suggests that there are a few plants with significantly higher heights in the data set, which pulls up the median, but the majority of plants have heights closer to the lower end, bringing down the mean. Therefore, the mean is lower and the median is higher in this data set.

7. Find the mean absolute deviation for the data set below. 180, 154, 210, 177, 194

183

76

15.2

14.6

The mean absolute deviation is a measure of how spread out the data set is. It is calculated by finding the average of the absolute differences between each data point and the mean of the data set. In this case, the mean of the data set is 183. To find the mean absolute deviation, we calculate the absolute difference between each data point and the mean, and then find the average of these differences. The absolute differences are: 3, 29, 27, 6, and 11. The average of these differences is 15.2, which is the mean absolute deviation for the given data set.

Explanation

The mean absolute deviation is a measure of how spread out the data set is. It is calculated by finding the average of the absolute differences between each data point and the mean of the data set. In this case, the mean of the data set is 183. To find the mean absolute deviation, we calculate the absolute difference between each data point and the mean, and then find the average of these differences. The absolute differences are: 3, 29, 27, 6, and 11. The average of these differences is 15.2, which is the mean absolute deviation for the given data set.

8. The min, Q1, Q2, Q3 and max of EOCT scores for 5 different math classes are listed below. Which class had the smallest middle 50% spread? Class A: 56, 71, 82, 89, 95 Class B: 64, 69, 78, 85, 92 Class C: 59, 67, 73, 80, 89 Class D: 64, 75, 83, 89, 94

Class A

Class B

Class C

Class D

The middle 50% spread refers to the range between the first quartile (Q1) and the third quartile (Q3). In this case, for Class A, the Q1 is 71 and the Q3 is 89. The difference between Q3 and Q1 is 18. For Class B, the difference is 17, for Class C it is 20, and for Class D it is 19. Therefore, Class A has the smallest middle 50% spread.

Explanation

The middle 50% spread refers to the range between the first quartile (Q1) and the third quartile (Q3). In this case, for Class A, the Q1 is 71 and the Q3 is 89. The difference between Q3 and Q1 is 18. For Class B, the difference is 17, for Class C it is 20, and for Class D it is 19. Therefore, Class A has the smallest middle 50% spread.