A student collects information about the number of school going children in a locality consisting of a hundred households. The data collected by him is
Class mark of a particular class is 9.5 and the class size is 6, then the class interval is
One of the sides of a frequency polygon is
Out of sixteen observations arranged in an ascending order, the 8th and 9th observations are 25 and 27. Then, the median is
Vihaan has marks of 92, 85, and 78 in three mathematics tests. In order to have an average of exactly 87 for the four math tests, he should obtain
To analyse the election results, the data is collected from a newspapers. The data thus collected is known as
The class size of a distribution is 25 and the first class interval is 200-224. Then, the class marks of first two class intervals are
class mark of 1st interval=212
class mark of 2nd interval=236.5
First Class interval=200-224
class mark=upper limit +lower limit divided by 2..
class mark of first interval is =200+224 divided by 2=212
Second class interval=224-249
class mark of second interval =224 +249 divided by 2
Which of the following is NOT a common measure of central tendency?
What is the median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84?
If the mean of x and 1/x is M, then the mean of x2 and
If the mean of x and 1/2 is M. then the mean of
x² and 1/x² is
Mean of x and 1/x is M. So
( x +1/x)/2=M
Squaring on both sides,
(x² + 1/x²+2*x*1/x)/4=M²
Dividing by 2 on both sides,
= (x²+1/x²)/2 =2 *(2m²-1)/2=2m²-1
=Mean of x² and 1/x²=(2m²-1)
Which of the following variables are discrete ? 1. Size of shoes, 2. Number of pages in a book, 3. Distance travelled by a train, 4. Time
In a frequency distribution, the mid-value of a class is 60.5 and the width of the class is 10. The lower limit of the class is
The mean for counting numbers through 100 is
Mode of a set of observations is the value which
The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers becomes 20. The excluded number is
For a given data, the difference between the maximum and minimum observation is known as its
The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is
Let x and y be the upper and lower class limit of frequency distribution.
Given, width of the class = 5
⇒ x-y= 5 …(i)
Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get
x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15.
Hence, the upper class limit of the highest class
=(Number of continuous classes x Class width + Lower class limit of the lowest class)
= 5 x 5+10 = 25+10=35
Hence,’the upper class limit of the highest class is 35.
After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.
Thus, the highest class is 30-35,
Hence, the upper limit of this class is 35.
The mean of first four prime numbers is
The mode of 4, 6, 7, 8, 12, 11, 13, 9, 13, 9, 7, 8, 9 is
The mean of five observations is 15. If the mean of first three observations is 14 and that of last three is 17, then the third observation is
⇒ It is given that Mean of 5 observations is 15.
∴ The sum of observations =15×5=75.
∴ It is given that mean of the first 3 observations is 14.
∴ The sum of first three observations =14×3=42.
⇒ Given that mean of the last 3 observations is 17.
∴ The sum of the last three observations =17×3=51
∴ The third observation =(42+51)−75
In an examination, ten students scored the following marks: 60, 58, 90, 51, 47, 81, 70, 95, 87, 99. The range of this data is
In a bar graph, 0.25 cm length of a bar represents 100 people. Then, the length of bar which represents 2000 people is
If, for the set of observations 4, 7, x, 8, 9, 10 the mean is 8, then x is equal to
A set of data consists of six numbers: 7, 8, 8, 9, 9 and x. The difference between the modes when x = 9 and x = 8 is
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. The mean of the given number is
Total numbers =50
Mean of numbers after subtracting 53 from each =3.5
Sum of numbers after subtracting 53 from each =3.5×50=175
Sum of the original numbers =175+53×50=2825
Mean of the original numbers =2825/50=56.5