List of top Mathematics Questions on Statistics

The mean of number 2, 4, 6, 8 is ____.

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are \( 2, 3, 5, 10, 11, 13, 15, 21 \), then the mean deviation about the median of all the 10 observations is:

If the observed frequencies exactly match expected frequencies, then what is \(\chi^2\) (Chi-square) value?

Mean and variance of an ungrouped data of 15 numbers are \(12\) and \(14\) respectively. For another ungrouped data of 15 numbers the mean and variance are \(14\) and \(10\) respectively. If the two data are combined, then the variance of the combined data of 30 numbers is

If all observations are increased by 5 their mean ____.

Mode of data 2, 3, 3, 5, 7 is ____.

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

The average age of 100 teachers in a college in 2000 was 50 years. In 2002, 20 teachers superannuated from their jobs, whose average age was 60 years. In 2005, 40 new teachers joined the college whose average age was 38 years. What was the average age of all the teachers in 2008?

Let \( \alpha \) and \( \beta \) respectively be the maximum and the minimum values of the function \( f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right) \). Then \( \alpha + 2\beta \) is equal to :

If the mean of first n natural numbers is \(\frac{6n}{11}\), then n is

Find the mean and the mode of the following frequency distribution :

The median and mode of a distribution are $25.2$ and $26.1$ respectively. The mean of the distribution is :

Assertion (A) : The mean of first 'n' natural numbers is \(\frac{n - 1}{2}\).
Reason (R) : The sum of first 'n' natural numbers is \(\frac{n(n + 1)}{2}\).

The mean and median of a frequency distribution are 43 and 43.4 respectively. The mode of the distribution is :

If the mean and mode of a data are 12 and 21 respectively, then its median is :

Find the mean and the mode for the following frequency distribution :

Assertion (A) : If the value of mode and mean for a distribution is 50 and 56 respectively, then the value of median is 54.
Reason (R) : Median = \(\frac{1}{3}\) (Mode - 2 Mean)

If \(\mu_1,\mu_2\) and \(\mu_3\) are the mean, median and mode respectively of a normal distribution, then \[ 3(\text{Median})-2(\text{Mode}) = ? \]

Assertion (A) : Median of a data is the value of \(\frac{N}{2}\), where N represents sum of all frequencies.
Reason (R) : Median divides the whole distribution in two equal parts.

The mean deviation from the median of the given frequency distribution is:

Mean deviation from the mean for the ungrouped data \[ 8,\,7,\,15,\,12,\,12,\,60,\,15,\,6,\,65,\,4,\,6,\,18 \] is

The range of the data 12, 25, 15, 18, 35, 11, 45, 33, 22, 10 is:

If the first quartile is 142 and the semi-interquartile range is 18, then the third quartile is:

The variance of 25 observations is 6. If each observation is multiplied by 3, then the new variance of the resulting observations is