Statistics

In an experiment with 15 observations on $x$, then following results were available: $\sum {{x^2}} = 2830$, $\sum x = 170$ One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is :

In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average marks of the girls?

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set :

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately :

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $\lambda$ and then each of them is decreased by 25, their mean remains the same. Then $\lambda$ is equal to :

The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then ${y \over x}$ is equal to

If for some x $\in$ R, the frequency distribution of the marks obtained by 20 students in a test is : .tg {border-collapse:collapse;border-spacing:0;} .tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;} .tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;} .tg .tg-s6z2{text-align:center} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-hgcj{font-weight:bold;text-align:center} Marks 2 3 5 7 Frequency (x + 1)² 2x - 5 x² - 3x x then the mean of the marks is

Let M denote the median of the following frequency distribution .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency 3 9 10 8 6 Then 20M is equal to :

Consider the following statements: (a) Mode can be computed from histogram (b) Median is not independent of change of scale (c) Variance is independent of change of origin and scale. Which of these is/are correct?

In a series of 2n observations, half of them equal $a$ and remaining half equal $–a$. If the standard deviation of the observations is 2, then $|a|$ equals

Let x₁, x₂,...........,x_n be n observations such that $\sum {x_i^2} = 400$ and $\sum {{x_i}} = 80$. Then a possible value of n among the following is

Suppose a population A has 100 observations 101, 102,........, 200, and another population B has 100 observations 151, 152,......., 250. If V_A and V_B represent the variances of the two populations, respectively, then ${{{V_A}} \over {{V_B}}}$ is

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

Statement - 1 : The variance of first n even natural numbers is ${{{n^2} - 1} \over 4}$ Statement - 2 : The sum of first n natural numbers is ${{n\left( {n + 1} \right)} \over 2}$ and the sum of squares of first n natural numbers is ${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$

If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then the d is equal to

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is