OP Malhotra Class 11 Maths Solutions Chapter 21 Measures of Dispersion Chapter Test

 

Question 1. Find the mean deviation from the mean for the following data : 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:
First, we find the mean (\( \bar{x} \)) of the given data points. We sum all the numbers and divide by the count of numbers, which is 10.
\( \bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} = 50 \)
Next, we calculate the absolute difference between each data point (\( x_i \)) and the mean (\( \bar{x} \)). This is \( |x_i - \bar{x}| \). We then sum these absolute differences.

The sum of absolute deviations is \( \Sigma |x_i - \bar{x}| = 84 \). Finally, we calculate the Mean Deviation (M.D.) about the mean.
M.D about mean \( = \frac{\Sigma|x_i-\bar{x}|}{n} = \frac{84}{10} = 8.4 \). The mean deviation shows how spread out the data points are, on average, from the mean.
In simple words: First, find the average of all numbers. Then, for each number, see how far it is from the average. Add up all these "distances" (ignoring negative signs) and divide by how many numbers you have. This gives you the mean deviation, which tells you how much the numbers typically vary from their average.

🎯 Exam Tip: Always calculate the mean accurately as the first step. Remember to take the absolute value of differences when finding mean deviation; do not consider negative signs.

 

Question 2. Find the mean deviation from the mean for the following data:
Answer:
To find the mean deviation for this frequency distribution, we first need to calculate the mean (\( \bar{x} \)). We use the formula \( \bar{x} = \frac{\Sigma f_ix_i}{\Sigma f_i} \).
Let's prepare a table to organize the calculations:

Using the direct method, we find the mean:
Mean \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{528}{66} = 8 \)
Now, we calculate the mean deviation from the mean:
M.D from mean \( = \frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i} = \frac{138}{66} = 2.09 \) (approximately).
In simple words: When you have numbers that appear multiple times (frequencies), first multiply each number by how many times it appears and sum them up. Then divide by the total count of all numbers to find the mean. After that, calculate how far each number is from this mean, multiply by its frequency, and sum them up again. Finally, divide this sum by the total count of numbers to get the mean deviation.

🎯 Exam Tip: For frequency distributions, remember to multiply the absolute deviations by their respective frequencies before summing them up. This ensures each observation's deviation is weighted correctly.

 

Question 3. Find the mean deviation for the mean for the following data:
Answer:
For grouped data (data in classes), we first find the mid-mark (\( x_i \)) for each class interval. Then, we calculate the mean (\( \bar{x} \)) using \( \bar{x} = \frac{\Sigma f_ix_i}{\Sigma f_i} \). After finding the mean, we can calculate the mean deviation using \( M.D. = \frac{\Sigma f_i|x_i-\bar{x}|}{\Sigma f_i} \).
Let's organize the data and calculations in a table. Here, \( \bar{x} = 27 \), as derived from the calculations:

By direct method, the mean is calculated as:
Mean \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1350}{50} = 27 \)
Thus, the Mean Deviation (M.D.) from the mean is:
M.D from mean \( = \frac{\Sigma f_i\left|x_i-27\right|}{\Sigma f_i} = \frac{512}{50} = 10.24 \).
In simple words: For data grouped into classes, find the middle value of each class. Use these middle values like individual data points and their frequencies to calculate the mean. Then, find the mean deviation using the same steps as for a normal frequency distribution. This helps understand the spread of data in categories.

🎯 Exam Tip: When dealing with class intervals, always calculate the mid-point (or mid-mark) for each class accurately. This value represents \( x_i \) in your calculations for mean and mean deviation.

 

Question 4. Find the mean deviation about the median for the following data : 11, 3, 8, 7, 5, 14, 10, 2, 9
Answer:
To find the mean deviation about the median, we first need to arrange the data in ascending order and find the median (Md).
Arranging the given data in ascending order, we have:
2, 3, 5, 7, 8, 9, 10, 11, 14
Here, the number of observations (\( n \)) is 9, which is an odd number.
The median (Md) for an odd number of observations is the \( \left(\frac{n+1}{2}\right) \)th observation.
Md \( = \left(\frac{9+1}{2}\right) \)th observation \( = 5 \)th observation
From the sorted data, the 5th observation is 8. So, Md \( = 8 \).
Now, we calculate the absolute deviation of each data point from the median (\( |x_i - Md| \)) and sum them up:

Finally, we calculate the Mean Deviation (M.D.) about the Median:
M.D about Median \( = \frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n} = \frac{27}{9} = 3 \).
In simple words: To find how much numbers vary from the middle value, first sort them from smallest to largest and find the exact middle number (the median). Then, for each number, find its distance from the median, ignoring if it's bigger or smaller. Add up all these distances and divide by how many numbers there are. This shows the average difference from the median.

🎯 Exam Tip: Always arrange the data in ascending or descending order before finding the median. Be careful when finding the median for an even number of observations; it's the average of the two middle values.

 

Question 5. Find the variance and standard deviation of the following data :
Answer:
To find the variance and standard deviation for this data with frequencies, we can use the assumed mean method. We'll choose an assumed mean (A), calculate deviations, and then apply the formulas.
Let's take Assumed Mean (A) = 98. The class interval (i) is not relevant here as the data is discrete.

Now, we can calculate the Variance:
Variance \( = \frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2 \)
\( = \frac{728}{22} - \left(\frac{44}{22}\right)^2 \)
\( = 33.090909 - (2)^2 \)
\( = 33.090909 - 4 \)
\( = 29.09 \) (approximately)

Next, we calculate the Standard Deviation (S.D.). Standard deviation is simply the square root of the variance.
S.D \( = \sqrt{\text{Variance}} = \sqrt{29.09} = 5.3935 \) (approximately).
In simple words: Variance tells you how much the numbers are spread out from their average, on average, by looking at squared differences. Standard deviation is the square root of variance, giving a more direct measure of this spread in the original units. Both help understand how consistent or varied a set of data is.

🎯 Exam Tip: Remember that variance is the mean of the squared deviations, and standard deviation is its square root. The assumed mean method simplifies calculations, especially with larger data values, by reducing the numbers you work with.

 

Question 6. Calculate the mean and variance after the following data :
Answer:
For grouped data with class intervals, we use the step deviation method to simplify calculations for the mean and variance. We first find the mid-marks, choose an assumed mean (A), and a common class interval (i).
Let's organize the calculations in a table:

Using the step deviation method, we calculate the mean:
Mean \( \bar{x} = A + \frac{\Sigma f_iu_i}{\Sigma f_i} \times i \)
\( = 105 + \frac{2}{30} \times 30 \)
\( = 105 + 2 \)
\( = 107 \)

Next, we calculate the variance using the step deviation formula:
Variance \( = \left[\frac{\Sigma f_iu_i^2}{\Sigma f_i} - \left(\frac{\Sigma f_iu_i}{\Sigma f_i}\right)^2\right] \times i^2 \)
\( = \left[\frac{76}{30} - \left(\frac{2}{30}\right)^2\right] \times 30^2 \)
\( = \left[2.5333 - \frac{4}{900}\right] \times 900 \)
\( = [2.5333 - 0.0044] \times 900 \)
\( = 2.5289 \times 900 \)
\( = 2276.01 \) (approximately 2276).
The calculation in the source is \( 76 \times 30 - 4 = 2276 \), which implies a simplified formula or rounding difference. We will follow the standard formula here.
In simple words: The step deviation method simplifies finding the mean and variance for data grouped into classes, especially when numbers are large. You choose a middle point (assumed mean) and divide deviations by the class size. This makes the numbers smaller and easier to calculate with, but you must remember to multiply back by the class size at the end.

🎯 Exam Tip: The step deviation method is efficient for grouped data with equal class intervals. Remember the \( i^2 \) term when calculating variance to correctly scale back from the \( u_i \) values.