Teachoo Quiz

Assertion Reasoning Quiz

Chapter 13 Class 11 Statistics | 7 questions | about 6 minutes

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Question 1 of 7
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Question 1 of 7
Assertion (A): The mean deviation about median for the dataset \(3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21\) is equal to \(5.27\).
Reason (R): For an odd number of observations \(n\), the median is given by the \((\frac{n+1}{2})^{\text{th}}\) observation when ordered.
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Question 2 of 7
Assertion (A): A standardized variable \(z_i = \frac{x_i - \bar{x}}{\sigma_x}\) created from dataset \(X\) has mean \(\bar{z} = 0\) and variance \(\sigma_z^2 = 1\).
Reason (R): Standardizing a variable converts raw observations into standard deviation units measured relative to the mean.
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Question 3 of 7
Assertion (A): The variance of the first \(n\) natural numbers is \(\frac{n^2 - 1}{12}\).
Reason (R): The sum of the first \(n\) natural numbers is given by \(\sum_{i=1}^n i = \frac{n(n+1)}{2}\).
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Question 4 of 7
Assertion (A): If the variance of \(X\) is \(\sigma_X^2 = 16\), then the variance of \(Y = -3X + 10\) is \(\sigma_Y^2 = -48\).
Reason (R): Variance is always non-negative because it is the average of squared deviations from the mean.
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Question 5 of 7
Assertion (A): Amplifying a signal voltage \(V\) by a factor of \(2.5\) and adding a \(+10\text{ mV}\) DC offset changes noise standard deviation from \(3\text{ mV}\) to \(7.5\text{ mV}\).
Reason (R): Adding a constant DC offset does not change standard deviation, while scaling by factor \(a = 2.5\) multiplies standard deviation by \(|a|\).
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Question 6 of 7
Assertion (A): For the observations \(\{12, 25, 18, 30, 15\}\), the Range is \(18\) and Mean Deviation about mean is \(6.0\).
Reason (R): The Range of a dataset is defined as the sum of its maximum and minimum values.
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Question 7 of 7
Assertion (A): Two groups of 50 students each have mean scores \(\bar{x}_1 = 60\) and \(\bar{x}_2 = 60\) with variances \(\sigma_1^2 = 4\) and \(\sigma_2^2 = 9\). The combined variance of all 100 students is \(6.5\).
Reason (R): When combining two datasets of equal size \(n_1 = n_2\), the combined arithmetic mean is equal to the average of their individual means, \(\bar{x} = \frac{\bar{x}_1 + \bar{x}_2}{2}\).
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