If the mean of the distribution is 2.6, then the value of y is Variatex|1|2|3|4|5| Frequency f of x|4|5|y|1|2|
We know that, Mean
Compute the median from the following table Marks obtained|No. of students| 0-10|2| 10-20|18| 20-30|30| 30-40|45| 40-50|35| 50-60|20| 60-70|6| 70-80|3|
None of these
Marks obtained|No. of students|Cumulative frequency|
Here n = 159, which is odd.
Median number, which is in the class 30-40 (see the row of cumulative frequency 95, which contains 80).
Hence median class is 30-40.
We have l = Lower limit of median class = 30
f = Frequency of median class = 45
C = Total of all frequencies preceding median class = 50
i = Width of class interval of median class = 10
Consider the following statements (1) Mode can be computed from histogram (2) Median is not independent of change of scale (3) Variance is independent of change of origin and scale Which of these is/are correct
(1), (2) and (3)
Only (1) and (2)
It is obvious.
The sum of two weights of an Atwood's machine is 16 lbs. The heavier weight descends 64 ft in 4 seconds. The value of weights in lbs are
Let the two weights be m1 and (lbs.)
Now, the distance moved by the heavier weight in 4 seconds from rest is 64 ft.
If acceleration = f, then ⇒
But ⇒ ⇒ ...(ii)
From (i) and (ii), lbs,
The following data gives the distribution of height of students Height (in cm)|160|150|152|161|156|154|155| Number of students|12|8|4|4|3|3|7| The median of the distribution is
Arranging the data in ascending order of magnitude, we obtain
Height (in cm)|150|152|154|155|156|160|161|
Number Of Students|8|4|3|7|3|12|4|
Cumulative frequency |8|12|15|22|25|37|41|
Here, total number of items is 41, i.e. an odd number. Hence, the median is th i.e. 21st item.
From cumulative frequency table, we find that median i.e. 21st item is 155.
(All items from 16 to 22nd are equal, each = 155)
A pie chart is to be drawn for representing the following data Items of expenditure|Number of Families| Education|150| Food and clothing|400| House rent|40| Electricity|250| Miscellaneous|160| The value of the central angle for food and clothing would be
Required angle for food and clothing
What is the standard deviation of the following series Measurements|0-10|10-20|20-30|30-40| Frequency|1|3|4|2|
Class Frequency yi ,
A = 25fiuifiui2|||
0-10|1|5|- 2|- 2|4|
10-20|3|15|- 1|- 3|3|
⇒ σ = 9
The S.D. of a variatexis σ. The S.D. of the variate where a, b, c are constant, is
Let i.e., i.e. , where
Thus, new S.D..
For the data The Karl Pearson's coefficient is
4 7 8 3 4|5 8 6 3 5|- 1 2 3 - 2 - 1|0 3 1 - 2 0|1 9 1 4 0|0 9 1 4 0|0 6 3 4 0 |
Three forces are acting at a point in a plane. The angles between and and and are 150° and 1200 respectively, then for equilibrium, forces P, Q, R are in the ratio
Clearly, the angle between P and R is
. By Lami's theorem,
A uniform beam of length 2a rests in equilibrium against a smooth vertical plane and over a smooth peg at a distance h from the plane. If θ be the inclination of the beam to the vertical, then sin3θ is
Let AB be a rod of length 2a and weight W. It rests against a smooth vertical wall at A and over peg C, at a distance h from the wall. The rod is in equilibrium under the following forces :
(i) The weight W at G
(ii) The reaction R at A
(iii) The reaction S at C perpendicular to AB.
Since the rod is in equilibrium. So, the three force are concurrent at O.
In ΔACK, we have, sin θ =
In ΔACO, we have, sin θ =
In ΔAGO, we have sin θ = ;