Math Problem Statement

The time (minutes) taken by students in class 11A and 11B to complete a question is given in a table. The problem asks to check the validity of several statistical statements, including the mean, standard deviation, and comparison of times between the two classes.

Solution

I see that you have uploaded a question in Vietnamese related to calculating statistical measures, such as the mean and standard deviation, from a data table. Let me break down the problem:

The table includes the time (in minutes) that students of two classes (11A and 11B) took to complete a question. You are asked to check the correctness of several statistical statements related to the data provided, including:

  • (a) The mean of the data for class 11A is 133.44.
  • (b) The standard deviation for class 11A is 11.77.
  • (c) The standard deviation for class 11B is 11.55.
  • (d) Comparing the standard deviation of the time taken by students in class 11A and 11B, the students in class 11A completed the task in less time, on average, than those in class 11B.

To clarify, I need to compute the values and check the validity of each statement. Let's start with calculating the mean and standard deviation for both classes.

First, let's extract the necessary data:

  • Class 11A:

    • Time: [0.5, 10.5, 20.5, 30.5, 40.5]
    • Number of students: [2, 10, 6, 4, 3]
  • Class 11B:

    • Time: [0.5, 10.5, 20.5, 30.5, 40.5]
    • Number of students: [3, 8, 10, 2, 4]

Let me perform the necessary calculations now. numpy as np

Data for class 11A

times_A = np.array([0.5, 10.5, 20.5, 30.5, 40.5]) students_A = np.array([2, 10, 6, 4, 3])

Data for class 11B

times_B = np.array([0.5, 10.5, 20.5, 30.5, 40.5]) students_B = np.array([3, 8, 10, 2, 4])

Calculating the mean for both classes

mean_A = np.average(times_A, weights=students_A) mean_B = np.average(times_B, weights=students_B)

Calculating the standard deviation for both classes

std_dev_A = np.sqrt(np.average((times_A - mean_A)**2, weights=students_A)) std_dev_B = np.sqrt(np.average((times_B - mean_B)**2, weights=students_B))

mean_A, std_dev_A, mean_B, std_dev_BHere are the results of the calculations:

  • Class 11A:

    • Mean time: 18.9 minutes
    • Standard deviation: 11.55 minutes
  • Class 11B:

    • Mean time: 19.02 minutes
    • Standard deviation: 11.77 minutes

Checking the given statements:

  • (a) The mean of class 11A is 133.44 – Incorrect. The mean time for class 11A is 18.9, not 133.44.
  • (b) The standard deviation of class 11A is 11.77 – Incorrect. The standard deviation for class 11A is 11.55, not 11.77.
  • (c) The standard deviation of class 11B is 11.55 – Incorrect. The standard deviation for class 11B is 11.77, not 11.55.
  • (d) Comparing the standard deviation: class 11A has a slightly lower standard deviation (11.55) than class 11B (11.77). Thus, class 11A is more consistent in their completion times than class 11B, so this statement is correct.

Would you like to go over any specific details or have further questions? Here's a follow-up list:

  1. How to interpret the meaning of standard deviation in the context of this data.
  2. How to calculate other statistical measures like variance.
  3. The role of weighting when calculating the mean and standard deviation.
  4. How to compare two datasets' consistency in more depth.
  5. How to handle outliers in this type of data analysis.

Tip: Always double-check the units and make sure you use the correct weightings when calculating weighted averages like the mean!

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Math Problem Analysis

Mathematical Concepts

Statistics
Mean
Standard Deviation
Weighted Average

Formulas

Mean = (Σx * w) / Σw
Standard Deviation = sqrt(Σ(w * (x - mean)^2) / Σw)

Theorems

Law of Large Numbers
Weighted Mean Theorem

Suitable Grade Level

Grades 9-12