Measures of Central Tendency and Dispersion PDF

## Measures of Central Tendency and Dispersion | Question | Answer A | Answer B | Answer C | Answer D | Correct | An Explanation | |---|---|---|---|---|---|---| | A research | Mean | Median | Mode | Range | B | The median is useful for understanding the central tendency of pulse rate increase, esp...

## Measures of Central Tendency and Dispersion | Question | Answer A | Answer B | Answer C | Answer D | Correct | An Explanation | |---|---|---|---|---|---|---| | A research | Mean | Median | Mode | Range | B | The median is useful for understanding the central tendency of pulse rate increase, especially if there are outliers in the data. | | In a survey | Mean | Median | Mode | Range | C | The mode is the most frequently occurring category, and in this case, it is 'Soccer.' | | A transport | Box Plot | Histogram | Pie Chart | Stem Plot | B | A histogram is suitable for displaying the distribution of continuous data like travel times and can show peak values. | | A marketing | Mean | Median | Mode | Range | B | The median would provide a more accurate central tendency for engagement times, as it is not affected by extreme values. | | A quality | Range | Standard Deviation | Interquartile Range | Mode | B | Standard deviation is a good measure of overall variability, especially when you have consistent values with no outliers. | | A university | The data is symmetrical | The data is bimodal | The data is skewed | The data has multiple peaks | B | The median and mean are close, it suggests that the data is likely symmetrical. | | A gym owner | Unimodal | Bimodal | Skewed to the left | Normally distributed | D | The data distribution suggest that it is bimodal or multimodal, indicating distinct groups of data within three standard deviations. | | A data analysis | Range | Interquartile Range | Standard Deviation | Mean | B | The interquartile range is less affected by extreme values and is a better measure of spread for skewed data. | | An engine | 4.7 to 5.3 | 4.8 to 5.2 | 4.9 to 5.1 | 5.0 to 5.4 | A | The values within three standard deviations: 5 ± 3 * 0.1 = 4.7 to 5.3 mm. | | A research | Box Plot | Scatter Plot | Histogram | Pie Chart | B | A scatter plot is used to visualize the relationship between two quantitative variables, such as study hours and grade. | | An insurance | Mean | Median | Mode | Range | B | A histogram is appropriate for visualizing the distribution of continuous data, such as flight prices, and identifying outliers. | | A travel company | Histogram | Bar Chart | Time Plot | Stem Plot | A | Association rule mining, such as market basket analysis, is used to identify products that are frequently bought together. | | A retailer | Scatter Plot | Association Rules | Box Plot | Histogram | B | A box plot allows for comparing the spread and variability of building heights across different neighborhoods. | | A city planner | Time Plot | Box Plot | Pie Chart | Stem Plot | B | A time plot will show how each client's weight changes over time and highlight trends in weight loss. | | A fitness center | Box Plot | Histogram | Time Plot | Bar Chart | C | A box plot is useful for comparing the spread and variability of GPAs across different academic programs. | | A university | Scatter Plot | Box Plot | Histogram | Bar Chart | B | A time plot is effective for identifying seasonal trends and patterns over a period of several years. | | A business | Box Plot | Scatter Plot | Time Plot | Bar Chart | C | The IQR is a robust measure of spread that minimizes the impact of extreme values, making it suitable for the data. | | A dietitian | Standard Deviation | Interquartile Range | Mean | Mode | C | The median is less affected by the skewness and provides a more accurate measure of central tendency for the data. | | A sociologist | Mean | Median | Mode | Range | B | A box plot will effectively highlight extreme wait times and show the overall distribution of patient wait time. | | A hospital | Histogram | Box Plot | Bar Chart | Stem Plot | B | The median is the best measure of central tendency when there is a wide range of scores with outliers. | | A game developer | Mean | Median | Mode | Standard Deviation | B | A box plot will help identify outliers and understand the distribution of pollutant concentrations in the river. | | A pharmaceutical | Mode | Median | Range | Standard Deviation | D | The median is less affected by extreme luxury home prices and would provide a better representation of typical prices, like the median house price. | | An environmental | Calculate | Generate | Create a pie chart | Use a scatter plot | B | A reaction time of 16 seconds is 3 standard deviations above the mean: 10 + 3 * 2 = 16 seconds. | | A real estate | Mean | Median | Mode | Range | B | 95% of the data falls within 2 standard deviations: 3.0 ± 2 * 0.5 = 2.0 to 4.0. | | A chemist | 12 second | 14 seconds | 16 seconds | 18 seconds | C | Using the IQR and a box plot is effective for identifying outliers, especially in skewed data. | | A school administrator | 2.5 to 3.5 | 2.0 to 4.0 | 1.5 to 4.5 | 3.0 to 4.0 | B | Q3 + 1.5 * IQR. First, find Q3: Q1 + IQR = 5 + 7 = 12. Upper boundary = 12 + 1.5 * 7 = 21.5. | | A data scientist | Calculate | Calculate | Calculate | Use a scatter plot | C | Q3 + 1.5 * IQR. First, find Q3: Q1 + IQR = 5 + 7 = 12. Upper boundary = 12 + 1.5 * 7 = 21.5. The upper boundary is ~22. Adding 1 to the upper boundary gives: 22 + 1 = 23. The difference is 23 - 21.5 = 1.5. | | An epidemiologist | 12 days | 15 days | 16.5 days | 18 days | C | The difference is 5 - 4, making the upper boundary 9. Adding 1 to the upper boundary gives: 9+1 = 10. The difference is 10 - 9 = 1. | | A sports analyst | 1 standard deviation | 2 standard deviations | 3 standard deviations | 0.5 standard deviation | B | 99.7% of the data falls within 3 standard deviations: 250 ± 3 * 20 = 180 to 320 grams. | | A research | 210 to 290 | 190 to 310 | 230 to 270 | 180 to 320 | D | Q1 - 1.5 * IQR = 140 - 1.5 * 20 = 130 cm. The lower boundary is 130. Adding 1 to the lower boundary gives: 130 + 1 = 131. The difference is 131 - 130 = 1. | | A teacher | 120 cm | 125 cm | 130 cm | 135 cm | C | Q3 + 1.5 * IQR = 5.2 - 1.5 * 0.4 = 4.8. The upper boundary is 4.8. Adding 1 to the upper boundary gives: 4.8 + 1 = 5.8. The difference is 5.8 - 5.2 = .6 | | A quality control | 4.8 mm | 5.0 mm | 5.2 mm | 5.4 mm | B | Q3 + 1.5 * IQR = 40 + 1.5 * 30 = 85. Since 100 is greater than 85, it is an outlier and is beyond the upper boundary. | | A dietitian | 1,750 to 2,100 | 1,500 to 2,100 | 2,250 to 2,100 | 1,800 to 2,100 | A | 68% of the data falls within one standard deviation of the mean: 1,000 ± 50 = 950 to 1,050 hours. | | A climate scientist | The data is symmetrical | The data is bimodal | The data is skewed | There are outliers in the data | B | The mean and IQR are generally good measures of central tendency for normally distributed data. | | A biologist | 1 standard deviation | 2 standard deviations | 3 standard deviations | 4 standard deviation | B | The standard deviation is appropriate for measuring the spread of ages and provides a clearer summary of the central spread of ages. | | An economist | Yes, it is a statistical outlier | No, it is not a statistical outlier | Only if it is greater than 1 standard deviation from the mean | Only if it is greater than 2 standard deviations from the mean | A | A box plot is effective for showing the normal distribution, and the standard deviation is appropriate for measuring the spread of ages. | | A statistic | 68% | 95% | 99.70% | 50% | A | The 68-95-99.7 rule, which is used for normal distribution, is effective for data which is not affected by extremely high-calorie meals. | | A geologist | Mean | Median | Mode | Range | B | The mean and standard deviation are appropriate measures of central tendency for the data since the data does not have outliers. | | An ecologist | Box Plot | Histogram | Scatter Plot | Bar Chart | B | For normally distributed data, the mean and standard deviation are appropriate measures of central tendency and the spread. | | A historian | Standard Deviation | Range | Interquartile Range | Mode | C | The IQR is less affected by extreme values and provides a robust measure of spread for the decay times. | | A nutritionist | Mean | Median | Mode | Range | B | The mean and standard deviation are adequate indicators of central tendency for the data. | | A marine biologist | Box Plot | Histogram | Pie Chart | Scatter Plot | A | The IQR is less affected by significantly longer response times and provides a better measure of the typical responses. | | A physicist | Range | Interquartile Range | Standard Deviation | Mode | B | The bimodal distribution, clearly representing the two groups: introverts and extroverts. | | An art historian | Mean | Median | Mode | Range | B | The median is appropriate for normally distributed data, and the mean is the best measure of central tendency. | | A psychologist | Mean | Median | Mode | Range | B | The median will be a better measure for skewed data, as it is not affected by the very tall statues. | | A sociologist | Histogram | Box Plot | Scatter Plot | Bar Chart | A | The distribution of pH levels and highlight any extreme acidity values. | | A space scientist | Median and Mode | Mean and Mode | Range and Standard Deviation | Mean and Interquartile Range | B | The distribution of genes with very high mutation rates and provides a clearer summary of the spread of mutation. | | A volcanologist | Range | Standard Deviation | Interquartile Range | Mode | B | A histogram is best to visualize the distribution and see if the ages are normally distributed, especially when there are outliers in the data. | | A wildlife | Scatter Plot | Histogram | Box Plot | Pie Chart | B | A box plot is especially useful for a small dataset of 30 values, as it preserves the exact data points while showing the distribution. Comparing the variability and spread of the data. |