FBQS
Page-22.27
Question 1:
The class-mark of the class 130-150 is ___________.
Answer 1:
The mid-value of a class is called the class mark.
The given class is 130–150.
∴ Class-mark of the given class
Thus, the class-mark of the class 130–150 is 140.
The class-mark of the class 130–150 is ___140___.
Question 2:
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is ________.
Answer 2:
Mid-value of the class = 10
Width of the class = 6
We know
Mid-value of the class = Lower limit of class + (Width of the class)
∴ 10 = Lower limit of the class + × 6
⇒ 10 = Lower limit of the class + 3
⇒ Lower limit of the class = 10 − 3 = 7
Thus, the lower limit of the class is 7.
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is ___7___.
Question 3:
In the class intervals 10-20, 20-30, the number 20 is included in the class ________.
Answer 3:
In the exclusive method of classification, the class intervals are so fixed that the upper limit of one class is the lower limit of the next class. Here, the upper limit of the class is not included in the class. It is included in the next class.
The given class intervals are 10–20, 20–30. Here, 20 is the upper limit of the class 10–20. So, it is not included in this class. The number 20 is included in the class interval 20–30.
In the class intervals 10–20, 20–30, the number 20 is included in the class __20–30__.
Question 4:
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 ________.
Answer 4:
We know
Width of the class = Upper class limit − Lower class limit
Lower class limit of the lowest class = 10
Width of the class = 5
∴ 5 = Upper class limit of the lowest class − 10
⇒ Upper class limit of the lowest class = 5 + 10 = 15
So, the lowest class is 10–15.
Now, the five continuous classes in the frequency distribution with class width 5 are 10–15, 15–20, 20–25, 25–30 and 30–35.
The highest class is 30–35. The upper class limit of the highest class is 35.
Thus, the upper class limit of the highest class is 35.
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 ___35___.
Question 5:
The class marks of a frequency distribution are given as follows : 15, 20, 25, ........ .The class corresponding to the class mark 20 is _________.
Answer 5:
The given class marks are 15, 20, 25,... . Here, the classes are uniformly spaced. So, the class width is the difference between any two consecutive class marks.
Class width = 20 − 15 = 5
If a is the class mark of a class interval of width h, then the lower and upper limits of the class interval are and , respectively.
Here, a = 20 and h = 5.
∴ Lower class limit of the class having class mark 20 =
Upper class limit of the class having class mark 20 =
So, the class corresponding to the class mark 20 is 17.5–22.5.
The class marks of a frequency distribution are given as follows : 15, 20, 25, ........ .The class corresponding to the class mark 20 is __17.5–22.5__.
Question 6:
In a frequency distribution, the mid value of a class is 42 and the width of the class is 10, then the class interval is ________.
Answer 6:
If a is the mid-value of a class interval of width h, then the lower and upper limits of the class interval are and , respectively.
Here, a = 42 and h = 10.
∴ Lower limit of the class interval =
Upper limit of the class interval =
Thus, the class interval is 37–47.
In a frequency distribution, the mid value of a class is 42 and the width of the class is 10, then the class interval is ___37–47___.
Question 7:
If l is the lower class limit of a class in a frequency distribution and m is the mid-point of the class, then the upper class limit of the class interval is ________.
Answer 7:
Mid-value of the class interval = m
Lower class limit of the class interval = l
We know
Mid-value of the class interval = (Lower class limit + Upper class limit)
∴ m = (l + Upper class limit)
⇒ l + Upper class limit = 2m
⇒ Upper class limit = 2m − l
Thus, the upper class limit of the class interval is 2m − l.
If l is the lower class limit of a class in a frequency distribution and m is the mid-point of the class, then the upper class limit of the class interval is ___2m − l___.
Question 8:
A grouped frequency distribution table with classes of equal size, using 63-72 (72 included) as one of the classes, is constructed for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44
The number of classes in the distribution will be __________.
Answer 8:
The given class is 63–72 (72 is included in the same class). So, the frequency distribution is to be formed by inclusive method.
Size of each class interval = 10 (72 is included in the given class)
Highest value of the observations = 112
Lowest value of the observations = 14
∴ Range of the data = Highest value of the observations − Lowest value of the observations = 112 − 14 = 98
Now,
= 9.8
So, the number of classes of equal size in the distribution will be 10. The class in the distribution are
13–22, 23–32, 33–42, 43–52, 53–62, 63–72, 73–82, 83–92, 93–102, 103–112
Thus, the number of classes in the distribution is 10.
The number of classes in the distribution will be ___10___.
Question 9:
A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included) as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 242, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236
The frequency of the class 310-330 is _________.
Answer 9:
The given class is 250–270 (270 is not included in the same class). So, the frequency distribution is to be formed by exclusive method. Here, the classes are continuous.
For the class 310–330, we take the observations between 310 and 330 including 310 and excluding 330.
The observations between 310 and 330 including 310 and excluding 330 in the given data are 310, 310, 320, 319, 318, 316.
So, the number of observations between 310 and 330 including 310 and excluding 330 in the given data is 6.
Thus, the frequency of the class 310–330 is 6.
The frequency of the class 310–330 is ____6____.
Question 10:
The range of the data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is __________.
Answer 10:
Range is the difference between the highest and lowest values of the given observations.
Here,
Highest value = 32
Lowest value = 6
∴ Range of the data = Highest value − Lowest value = 32 − 6 = 26
Thus, the range of the data is 26.
The range of the data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is ____26____.
Question 11:
The marks obtained by 17 students in a Mathematics test (out of 100) are given below:
91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49.
The range of the data is _________.
Answer 11:
Range is the difference between the highest and lowest values of the given observations.
Here,
Maximum marks obtained = 100
Minimum marks obtained = 46
∴ Range of the data = Maximum marks obtained − Minimum marks obtained = 100 − 46 = 54
Thus, the range of the data is 54.
The marks obtained by 17 students in a Mathematics test (out of 100) are given below:
91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49.
The range of the data is ___54____.
Question 12:
The class intervals in a frequency distribution are:
40 – 44, 45 – 49, 50 – 54,........ If the frequency distribution, is converted into a continuous frequency distribution then the class intervals are _________.
Answer 12:
If a–b is a class in inclusive method, then in exclusive method the corresponding class is –, where h = (Lower limit of a class − Upper limit of previous class).
In the given class intervals, the difference between the lower limit of a class and upper limit of the previous class is 1. Here, h = = 0.5.
So, we subtract 0.5 from the lower limit of each class and add 0.5 in the upper limit of each class to make it continuous.
For the class 40–44, the corresponding class is (40 − 0.5)–(44 + 0.5) i.e. 39.5–44.5
For the class 45–49, the corresponding class is (45 − 0.5)–(49 + 0.5) i.e. 44.5–49.5
For the class 50–54, the corresponding class is (50 − 0.5)–(54 + 0.5) i.e. 49.5–54.5 and so on
Thus, the class intervals in the continuous frequency distribution are 39.5–44.5, 44.5–49.5, 49.5–54.5, ... .
The class intervals in a frequency distribution are: 40 – 44, 45 – 49, 50 – 54,........ If the frequency distribution, is converted into a continuous frequency distribution then the class intervals are ___39.5–44.5, 44.5–49.5, 49.5–54.5, ...___.
Question 13:
Let m be the mid-point and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is __________.
Answer 13:
Mid-value of the class = m
Upper limit of the class = u
We know
Mid-value of the class = (Lower limit of the class + Upper limit of the class)
∴ m = (Lower limit of the class + u) (Given)
⇒ Lower limit of the class + u = 2m
⇒ Lower limit of the class = 2m − u
Thus, the lower limit of the class is 2m − u.
Let m be the mid-point and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is ___2m − u____.
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