Problem 47
Question
Suppose a population \(A\) has 100 observations 101 , \(102, \ldots, 200\), and another population \(B\) has 100 observations \(151,152, \ldots, 250\). If \(V_{A}\) and \(V_{B}\) represent the variances of the two populations, respectively, then \(\frac{V_{A}}{V_{B}}\) is [2006] (A) 1 (B) \(9 / 4\) (C) \(4 / 9\) (D) \(2 / 3\)
Step-by-Step Solution
Verified Answer
Option (A): 1.
1Step 1: Understand the definition of variance
The variance of a population is calculated as the average of the squared differences from the mean. For a set of numbers, it's calculated as follows: \[ V = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \] where \(\mu\) is the mean of the data set and \(n\) is the number of observations.
2Step 2: Calculate the mean of population A
Population A has observations from 101 to 200. The mean \(\mu_A\) is calculated as: \[ \mu_A = \frac{101 + 102 + \ldots + 200}{100} = \frac{\text{sum of first 200} - \text{sum of first 100}}{100} = \frac{10100 - 5050}{100} = 150.5 \]
3Step 3: Calculate the mean of population B
Population B has observations from 151 to 250. The mean \(\mu_B\) is calculated as: \[ \mu_B = \frac{151 + 152 + \ldots + 250}{100} = \frac{\text{sum of first 250} - \text{sum of first 150}}{100} = \frac{31375 - 11325}{100} = 200.5 \]
4Step 4: Calculate variance of population A
The variance \(V_A\) is calculated as: \[ V_A = \frac{1}{100} \sum_{i=0}^{99} (101 + i - 150.5)^2 \] Simplifying, you find that this results in \(V_A = \frac{1}{100} (\sum_{i=0}^{99} (i + 49.5)^2)\). Using properties of sums, this simplifies to \(V_A = 841.667\).
5Step 5: Calculate variance of population B
The variance \(V_B\) is calculated as: \[ V_B = \frac{1}{100} \sum_{i=0}^{99} (151 + i - 200.5)^2 \] This results in \(V_B = \frac{1}{100} (\sum_{i=0}^{99} (i + 49.5)^2)\). Simplifying further using the properties of sums, it returns \(V_B = 841.667\) similar to \(V_A\).
6Step 6: Calculate the ratio of the variances
Now, calculate the ratio \(\frac{V_A}{V_B}\):\[ \frac{V_A}{V_B} = \frac{841.667}{841.667} = 1 \] So, the ratio of the variances is 1.
Key Concepts
Population MeanSum of SeriesVariance CalculationObservations
Population Mean
The population mean is a fundamental concept in statistics. It's essentially the average value of a data set and gives a central value around which the data tends to cluster. To calculate the population mean, sum all the observations and then divide by the number of observations.
For instance, in population A, which includes numbers from 101 to 200, we added up all these numbers and divided by 100 since there are 100 observations. This gave us a mean (\(\mu_A\)) of 150.5. Similarly, for population B, which ranges from 151 to 250, we also added the numbers and found the mean to be 200.5. This mean is crucial because it serves as the reference point for calculating variance.
For instance, in population A, which includes numbers from 101 to 200, we added up all these numbers and divided by 100 since there are 100 observations. This gave us a mean (\(\mu_A\)) of 150.5. Similarly, for population B, which ranges from 151 to 250, we also added the numbers and found the mean to be 200.5. This mean is crucial because it serves as the reference point for calculating variance.
Sum of Series
To find the population mean, we need to calculate the sum of a series. The sum of a series refers to the total obtained by adding a sequence of numbers.
For a series like 101 to 200 or 151 to 250, a formula exists that helps compute the sum efficiently: \[\text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term})\]This formula allows us to quickly determine the sum of numbers without adding each individually. For instance, the sum of numbers from 101 to 200 is 15,050, as calculated using this formula.
For a series like 101 to 200 or 151 to 250, a formula exists that helps compute the sum efficiently: \[\text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term})\]This formula allows us to quickly determine the sum of numbers without adding each individually. For instance, the sum of numbers from 101 to 200 is 15,050, as calculated using this formula.
- This approach simplifies calculations, especially when dealing with large datasets.
Variance Calculation
Variance measures how spread out the numbers in a data set are. It tells us the degree to which each number differs from the mean. A low variance indicates that the numbers are close to the mean, while a high variance means they are spread out over a wider range.
To calculate variance, subtract the mean from each observation to find the deviation, then square this deviation, sum all the squared deviations, and finally average them by dividing by the number of observations: \[V = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2\]For the populations in the exercise, both variances were calculated to be 841.667.
To calculate variance, subtract the mean from each observation to find the deviation, then square this deviation, sum all the squared deviations, and finally average them by dividing by the number of observations: \[V = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2\]For the populations in the exercise, both variances were calculated to be 841.667.
- This calculation involves using distribution properties, simplifying mathematically to obtain the variance.
- By comparing variances, we can understand the relative spread of data sets, as we did in determining the ratio \( \frac{V_A}{V_B} = 1 \).
Observations
In statistics, observations are the individual data points within a dataset. They form the basis for statistical analysis and calculations. Each observation is a specific value that, when combined with others, provides insights into the population's characteristics.
In the exercise, for populations A and B, each population consists of 100 observations, such as 101 or 151, respectively. These values mattered because:
In the exercise, for populations A and B, each population consists of 100 observations, such as 101 or 151, respectively. These values mattered because:
- They were needed to find the population mean.
- They were the elements over which variance was calculated.
Other exercises in this chapter
Problem 45
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [2005] (A) \(22.0\) (B) \(20.5\) (C) \(25.5\) (D)
View solution Problem 46
Let \(x_{1}, x 2, \ldots, x_{n}\) be \(n\) observations such that \(\sum x_{i}^{2}=400\) and \(\sum x_{i}=80 .\) Then a possible value of \(n\) among the follow
View solution Problem 48
The average marks of boys in a class is 52 and that of girls is 42 . The average marks of boys and girls combined is 50 . The percentage of boys in the class is
View solution Problem 49
The mean of the numbers \(a, b, 8,5,10\) is 6 and the variance is \(6.80 .\) Then which one of the following gives possible values of \(a\) and \(b ?\) (A) \(a=
View solution