Statistics

Solution

The mean, or average, is a measure of central tendency that represents the sum of all observations divided by the number of observations. For grouped data, the mean is calculated using the formula: mean = (Σf_i x_i) / Σf_i, where f_i is the frequency of the ith class and x_i is the class mark. For example, if we have a frequency distribution with classes 10-20, 20-30, etc., we first find the class mark (midpoint) of each class, multiply each class mark by its frequency, sum these products, and then divide by the total number of observations. This method is known as the direct method. Another method is the assumed mean method, which simplifies calculations by assuming a mean and adjusting the final result accordingly.

Solution

The mode is the value that appears most frequently in a data set. For grouped data, the mode is found using the formula: mode = l + ((f1 - f0) / (2f1 - f0 - f2)) * h, where l is the lower limit of the modal class, f1 is the frequency of the modal class, f0 is the frequency of the class preceding the modal class, f2 is the frequency of the class succeeding the modal class, and h is the class size. To find the mode, first identify the modal class (the class with the highest frequency), then apply the formula. For example, if the modal class is 30-40 with a frequency of 15, the preceding class has a frequency of 10, and the succeeding class has a frequency of 12, and the class size is 10, the mode would be calculated as 30 + ((15 - 10) / (30 - 10 - 12)) * 10 = 30 + (5 / 8) * 10 ≈ 36.25.

Solution

The median is the middle value in a data set when the observations are arranged in ascending or descending order. For grouped data, the median is calculated using the formula: median = l + ((n/2 - cf) / f) * h, where l is the lower limit of the median class, n is the total number of observations, cf is the cumulative frequency of the class preceding the median class, f is the frequency of the median class, and h is the class size. To find the median, first determine the median class (the class where the n/2th observation lies), then apply the formula. For example, if the median class is 40-50 with a cumulative frequency of 25 up to the preceding class, a frequency of 10 in the median class, and a class size of 10, and there are 50 observations in total, the median would be 40 + ((25 - 25) / 10) * 10 = 40.

Solution

Cumulative frequency is the sum of the frequencies of all classes up to a certain class. It is used to construct an ogive, which is a graph that shows the cumulative frequency distribution. To construct an ogive, plot the upper class limits on the x-axis and the corresponding cumulative frequencies on the y-axis, then connect the points with a smooth curve. For example, if we have classes 10-20, 20-30, etc., with cumulative frequencies of 5, 15, 30, etc., we would plot points at (20,5), (30,15), (40,30), etc., and connect them to form the ogive. The ogive can be used to find the median and other percentiles by locating the corresponding values on the graph.

Solution

The mean, median, and mode are all measures of central tendency, but they have different properties and uses. The mean is the arithmetic average and is influenced by every value in the data set, making it sensitive to outliers. The median is the middle value and is less affected by outliers, making it better for skewed distributions. The mode is the most frequent value and is useful for categorical data. For example, the mean is appropriate for normally distributed data like test scores, the median is better for income data which is often skewed, and the mode is best for identifying the most common category, such as the most popular shoe size. Each measure provides different insights into the data and should be chosen based on the data's characteristics and the analysis's goals.

Solution

The class mark is the midpoint of a class interval and is calculated as (lower limit + upper limit) / 2. It represents the central value of the class and is used in calculating the mean for grouped data by multiplying each class mark by its frequency and summing these products before dividing by the total number of observations. For example, for the class interval 20-30, the class mark is (20 + 30) / 2 = 25. If this class has a frequency of 5, it contributes 25 * 5 = 125 to the sum of products. The class mark simplifies the calculation of the mean by providing a single representative value for each class.

Solution

The assumed mean method is a simplified way to calculate the mean of grouped data by assuming a mean (a) and adjusting the final result. The steps are: 1) Choose an assumed mean (a), usually the class mark of the central class. 2) Calculate the deviation (d_i) of each class mark from the assumed mean, where d_i = x_i - a. 3) Multiply each deviation by its frequency (f_i * d_i). 4) Sum these products and divide by the total number of observations to get the mean deviation. 5) Add the assumed mean to the mean deviation to get the actual mean. For example, if the assumed mean is 50 and the sum of f_i * d_i is 100 with a total of 20 observations, the mean deviation is 100 / 20 = 5, and the actual mean is 50 + 5 = 55.

Solution

The step-deviation method further simplifies the assumed mean method by scaling the deviations by the class size (h). The steps are: 1) Choose an assumed mean (a) and class size (h). 2) Calculate the scaled deviation (u_i) for each class mark, where u_i = (x_i - a) / h. 3) Multiply each u_i by its frequency (f_i * u_i). 4) Sum these products and divide by the total number of observations to get the mean scaled deviation. 5) Multiply the mean scaled deviation by h and add the assumed mean to get the actual mean. For example, if a = 50, h = 10, and the sum of f_i * u_i is 20 with a total of 100 observations, the mean scaled deviation is 20 / 100 = 0.2, and the actual mean is 50 + (0.2 * 10) = 52. This method reduces the size of numbers involved, making calculations easier.

Solution

To determine the median class in a grouped frequency distribution, follow these steps: 1) Calculate the total number of observations (n). 2) Find n/2 to locate the median position. 3) Construct a cumulative frequency table to identify the class where the n/2th observation lies. This class is the median class. For example, if n = 50, the median position is at 25. If the cumulative frequency reaches 20 up to the class 30-40 and 35 up to the class 40-50, the 25th observation falls in the 40-50 class, making it the median class. The median is then calculated using the median formula for grouped data within this class.

Solution

Measures of central tendency (mean, median, mode) are essential in statistics as they summarize a large data set with a single value representing the center of the data. They help in understanding the typical value around which data points cluster. For example, the mean provides an average value useful for comparing different data sets, such as average rainfall in different months. The median gives the middle value, useful for understanding the distribution's center in skewed data, like household income. The mode identifies the most frequent value, helpful in inventory management to stock the most demanded item. These measures simplify data interpretation, facilitate comparisons, and support decision-making by providing a clear summary of the data's central tendency.

Solution

Mean is the average of all observations, median is the middle value when data is arranged in order, and mode is the most frequently occurring value. For example, in a dataset of test scores: 10, 20, 20, 30, the mean is 20, median is 20, and mode is 20. Mean is affected by extreme values, median is best for skewed distributions, and mode is useful for categorical data.

Solution

First, find the class marks (midpoints) of each class: 1, 3, 5, 7, 9, 11, 13. Multiply each class mark by its frequency and sum the products: (1*1)+(3*2)+(5*1)+(7*5)+(9*6)+(11*2)+(13*3) = 1+6+5+35+54+22+39 = 162. Divide by total number of houses (20): 162/20 = 8.1 plants per house.

Solution

The assumed mean method simplifies calculations by subtracting a guessed mean (a) from each observation, reducing the size of numbers. The step-deviation method further simplifies by dividing these differences by a common factor (h), making calculations with large numbers easier. Assumed mean is efficient when data is not too large, while step-deviation is better for large datasets with a common class size.

Solution

First, ensure cumulative frequencies are calculated. The median class is 500-600. Using the median formula: 525 = 500 + [(50 - (36 + x))/20]*100. Solving gives x = 9. Since total frequency is 100, x + y = 24 → y = 15.

Solution

The modal class is the class with the highest frequency. However, the mode may not be representative if the data is multimodal (multiple modes) or if the highest frequency is not significantly higher than others, leading to ambiguity in central tendency.

Solution

Convert to class intervals and frequencies: below 140 (4), 140-145 (7),...,160-165 (5). Median position is 25.5, falling in 145-150. Using the median formula: Median = 145 + [(25.5 - 11)/18]*5 ≈ 149.03 cm.

Solution

The empirical relationship is Mode = 3*Median - 2*Mean. If mean > median > mode, the data is right-skewed. If mean < median < mode, it's left-skewed. This helps in understanding the distribution's symmetry.

Solution

Choose a = 550 (midpoint), h = 20. Calculate u_i = (x_i - 550)/20 for each class. Multiply u_i by frequencies and sum: (-2.5*12)+(-1.5*14)+...+ (2.5*10) = -30 -21 -8 +0 +25 = -34. Mean = 550 + (-34/50)*20 = 550 - 13.6 = 536.4.

Solution

Median position is 200, within 3000-3500 class. Using the median formula: Median = 3000 + [(200 - 130)/86]*500 ≈ 3000 + 406.98 ≈ 3406.98 hours.

Solution

Median position is 50, within 7-10 class. Median = 7 + [(50 - 36)/40]*3 ≈ 8.05 letters. Mean could be affected by extreme values, and mode might not be unique or representative if multiple classes have similar high frequencies.

Solution

When class intervals are of unequal widths, the calculation of mean, median, and mode becomes more complex. For the mean, the assumption that frequencies are centered at the mid-point may not hold, leading to potential inaccuracies. The median requires adjusting the formula to account for varying class sizes, and the mode's formula may not directly apply, necessitating a more nuanced approach. Examples include histograms with varying bin sizes where the area of bars represents frequency.

Solution

The new mean will be 105. When each observation is multiplied by 2, the mean also gets multiplied by 2, becoming 100. Adding 5 to each observation increases the mean by 5, resulting in 105. This is derived from the properties of linear transformations on data sets.

Solution

In a country with high income inequality, the mean can be skewed by extremely high incomes, making it a less representative measure. The median, being the middle value, is less affected by outliers and provides a better central tendency measure. The mode, representing the most frequent income, may not reflect the overall economic condition. Choosing the median avoids the distortion caused by extreme values, offering a more accurate picture of the typical income.

Solution

To find the median from a cumulative frequency distribution: 1) Calculate the total number of observations (N). 2) Determine the median position as (N+1)/2 for odd N or N/2 and (N/2)+1 for even N. 3) Locate the class interval containing the median position using the cumulative frequencies. 4) Apply the median formula for grouped data: Median = L + [(N/2 - CF)/f] * h, where L is the lower limit of the median class, CF is the cumulative frequency before the median class, f is the frequency of the median class, and h is the class width. This method efficiently locates the median without enumerating all data points.

Solution

The mode is indeed suitable for categorical data as it identifies the most frequent category, which is meaningful when data is non-numeric (e.g., favorite colors). However, it may not be the 'best' if the data has multiple modes (multimodal distribution) or if the most frequent category is not significantly more common than others, leading to ambiguity. For example, in a survey of preferred transportation modes with nearly equal votes for 'bus' and 'train,' the mode doesn't provide a clear central tendency.

Solution

To determine the relationship, one could use correlation analysis, specifically Pearson's correlation coefficient, which measures the linear relationship between two variables. A positive value indicates that as hours studied increase, so do exam scores. Limitations include its sensitivity to outliers and inability to capture non-linear relationships. Additionally, correlation does not imply causation; other confounding variables may influence the results.

Solution

A frequency polygon can compare two data sets by plotting their frequencies on the same graph, using different lines for each set. This allows for easy visual comparison of distributions. Advantages over histograms include clarity in overlapping data and the ability to display multiple distributions without clutter. For example, comparing test scores of two classes is more straightforward with polygons than overlapping histograms.

Solution

In real estate, home prices in a neighborhood can vary widely, with a few luxury homes skewing the average price upwards. The median price, being the middle value, is less affected by these outliers and better represents what a typical home costs. For instance, if most homes are priced around $300,000 but a few are over $3 million, the mean would be misleadingly high, whereas the median remains representative of the majority.

Solution

Missing values can skew the mean if not handled, as the calculation assumes all data points are present. The median is more robust but may shift if missing values are not random. The mode might become unreliable if the most frequent value is missing. Mitigation strategies include imputing missing values (using mean/median/mode), using models to predict missing data, or analyzing complete cases only, each with its own assumptions and limitations.

Solution

A dataset where mean, median, and mode are equal is perfectly symmetrical with a single peak (unimodal). An example is the normal distribution, where data is evenly distributed around the center. This symmetry implies no skewness, making all three measures equally representative. In practice, such datasets simplify analysis as any measure of central tendency provides the same insight, but they are rare in real-world scenarios where skewness and outliers are common.