OP Malhotra Class 11 Maths Solutions Chapter 20 Measures of Central Tendency Chapter Test

 

Question 1. The weights of 50 apples were recorded as given below. Calculate the mean weight to the nearest gram, by step deviation method.

Answer: To find the mean weight using the step deviation method, we first create a frequency distribution table with mid-marks, deviations, and step deviations. The class size \( h \) for this data is 5. We assume an arbitrary mean \( A = 97.5 \).Now, we use the step deviation formula to calculate the mean. \( \bar{x} = A + \frac{\Sigma f_i u_i}{\Sigma f_i} \times h \)
\( \implies \) \( \bar{x} = 97.5 + \frac{(-16)}{50} \times 5 \)
\( \implies \) \( \bar{x} = 97.5 + (-0.32) \times 5 \)
\( \implies \) \( \bar{x} = 97.5 - 1.6 \)
\( \implies \) \( \bar{x} = 95.9 \) So, the mean weight of the apples is 95.9 grams. The step deviation method simplifies calculations for larger datasets.In simple words: We find the middle value for each weight group. Then, we use a special formula that helps us find the average weight by working with smaller numbers first. This gives us the final average weight of all the apples.

🎯 Exam Tip: Always remember to correctly identify the assumed mean \(A\) and class interval \(h\) (or \(i\)) for the step deviation method. Double-check your \(f_i u_i\) calculations to avoid errors in the sum.

 

Question 2. Find the value of p if the mean of the following distribution is 7.5 :

Answer: We need to find the missing frequency \( p \). First, we set up a table to calculate \( \Sigma f_i \) and \( \Sigma f_i x_i \). The mid-mark (\( x_i \)) for each class is the average of its lower and upper limits.Using the direct method, the formula for the mean is: \( \text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i} \) We are given that the mean is 7.5. So, we can set up the equation:
\( \implies \) \( 7.5 = \frac{303 + 9p}{41 + p} \) Now, we solve for \( p \):
\( \implies \) \( 7.5 (41 + p) = 303 + 9p \)
\( \implies \) \( 307.5 + 7.5p = 303 + 9p \)
\( \implies \) \( 307.5 - 303 = 9p - 7.5p \)
\( \implies \) \( 4.5 = 1.5p \)
\( \implies \) \( p = \frac{4.5}{1.5} \)
\( \implies \) \( p = 3 \) The value of the missing frequency \( p \) is 3. This method is effective for finding unknowns when the mean is given.In simple words: We list the middle numbers for each group and multiply them by their frequencies. We add up all these numbers, and also add up all the frequencies. Since we know the average, we use it in a formula to find the missing count, which turns out to be 3.

🎯 Exam Tip: When a frequency is unknown, represent it with a variable like 'p' and carry it through your summation calculations. Remember to set up and solve the equation for the mean carefully.

 

Question 3. The mean of the following frequency distribution is 57.6 and the number of observations is 50. Find the missing frequencies f₁ and f2.

Answer: We need to find the missing frequencies, denoted as \( f_1 \) and \( f_2 \) (instead of 'A' as in the question table). Let the missing frequencies be \( f_1 \) and \( f_2 \). We will use the step deviation method. Let the assumed mean \( A = 50 \) and the class interval \( C = 20 \).We are given that the total number of observations is 50. So, \( \Sigma f_i = 50 \)
\( \implies \) \( 32 + f_1 + f_2 = 50 \)
\( \implies \) \( f_1 + f_2 = 50 - 32 \)
\( \implies \) \( f_1 + f_2 = 18 \) (Equation 1) Now, using the step deviation method, the mean is given by: \( \text{Mean} = A + \frac{\Sigma f_i u_i}{\Sigma f_i} \times C \) We are given the mean is 57.6, \( A = 50 \), \( \Sigma f_i = 50 \), and \( C = 20 \).
\( \implies \) \( 57.6 = 50 + \frac{17 - f_1 + f_2}{50} \times 20 \)
\( \implies \) \( 57.6 - 50 = \frac{(17 - f_1 + f_2) \times 20}{50} \)
\( \implies \) \( 7.6 = \frac{17 - f_1 + f_2}{5} \times 2 \) Multiply both sides by 5:
\( \implies \) \( 7.6 \times 5 = (17 - f_1 + f_2) \times 2 \)
\( \implies \) \( 38 = 2 (17 - f_1 + f_2) \) Divide both sides by 2:
\( \implies \) \( 19 = 17 - f_1 + f_2 \)
\( \implies \) \( 19 - 17 = -f_1 + f_2 \)
\( \implies \) \( 2 = -f_1 + f_2 \)
\( \implies \) \( f_2 - f_1 = 2 \) (Equation 2) Now we have a system of two linear equations: 1) \( f_1 + f_2 = 18 \) 2) \( f_2 - f_1 = 2 \) Add Equation 1 and Equation 2: \( (f_1 + f_2) + (f_2 - f_1) = 18 + 2 \)
\( \implies \) \( 2f_2 = 20 \)
\( \implies \) \( f_2 = 10 \) Substitute the value of \( f_2 \) into Equation 1: \( f_1 + 10 = 18 \)
\( \implies \) \( f_1 = 18 - 10 \)
\( \implies \) \( f_1 = 8 \) So, the missing frequencies are \( f_1 = 8 \) and \( f_2 = 10 \). This two-step process, combining total frequency and the mean formula, is standard for finding two unknowns.In simple words: We are given the total number of items and the average. We use these two facts to create two math puzzles (equations) with the two missing numbers. By solving these puzzles, we find that the first missing number is 8 and the second is 10.

🎯 Exam Tip: When solving for two missing frequencies, you will always need two equations: one from the sum of frequencies and another from the mean formula. Be careful with signs when rearranging the equations.