Rd Sharma 2021 for Class 9 Maths Chapter 22 - Tabular Representation Of Statistical Data

Answer:

(i) In singular form statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data.

(ii) In plural form statistics means numerical facts or observations collected with definite purpose.

For examples, the income and expenditure of persons in a particular locality, number of males and females in a particular town are statistics.

Answer:

The plural form statistics has the simplest structure and the singular form statistics has many components. There is only structural difference between singular and plural form statistics. Some of the characteristics of a statistics are

1. Statistics is a collection of observations. So, clearly a single observation cannot form a statistics.

2. Statistics are collected with definite purpose.

3. Statistics are comparable and classified into various types depending on their properties.

4. Statistics are expressed quantitatively and not qualitatively. 

Answer:

(i) When an investigator collects data himself with a definite plan or designs in his (her) mind, it is called primary data.

(ii) Data which are not originally collected rather obtained from published or unpublished sources are known as secondary data.

Answer:

To study the features of a collected data, the data must be arranged in a condensed form. There are a number of ways to arrange the data in condensed form, namely,

1. Serial order or alphabetical order

2. Ascending order

3. Descending order

But, if the number of observations is large, then arranging data in ascending or descending or serial order is a tedious job and it does not tell us much except perhaps the minimum(s) and maximum(s) of data. So to make it easily understandable and clear we condense the data into groups or table form.

Answer:

(i) A name which takes different values is called variates. For example, the mark obtained by students of class IX in mathematics is variates.

(ii) In a grouped data, the groups are called class-intervals. For example, 0-5, 5-10, 10-15… are class intervals.

(iii) The difference between the true upper limit and the true lower limit of a class is called its class size. For example, the class size of the class-interval 10-15 is

(iv) The mid value of a class is called the class mark. For example, the mid value of the class 10-15 is

(v) The number of observation falling in a particular class is called the frequency of that class or class frequency. For example, if the number of students obtaining marks 60-70 in a particular subject is 60, then the frequency of the class 60-70 is 60.

(vi) Class limits are the boundaries of a class. The left boundary of a class is called the lower limit and the right boundary of a class is called the upper limit. For example, for the class interval 60-70 the lower limit is 60 and the upper limit is 70.

(vii) The class limits of a continuous grouped frequency distribution are called true class limits. For example, 5-10, 10-15, 15-20, 20-25, 25-30 are continuous class intervals, then the true lower and upper limits (class limits) of the class 15-20 are 15 and 20 respectively. If the classes are not continuous, then adjust the class intervals to form continuous grouped class intervals.

Answer:

After arranging in ascending order, the ages of the students are 7 years 8 months, 7 years 10 months, 8 years 4 months, 8 years 6 months, 8 years 7 months, 8 years 8 months, 8 years 11 months, 9 years, 9 years 2 months, and 9 years 3 months.

(i) The lowest age is 7 years 8 months.

(ii) The highest age is 9 years 3 months.

(iii) The range of the ages is

Answer:

(i) After arranging in ascending order, the pocket money’s in Rs. are 25, 30, 40, 45, 45 and 50. The highest pocket money is Rs 50.

(ii) The lowest pocket money is Rs 25.

(iii) The range of the pocket money’s is

(iv) The given data (pocket money in Rs.) in ascending order is 25, 30, 40, 45, 45 and 50.

Answer:

(i) The given classes are 0-4, 5-9 and 10-14. The classes can be written in continuous form as (-0.5)-4.5, 4.5-9.5, and 9.5-14.5. The upper and lower limits of the first class are 4.5 and (-0.5) respectively. Hence, the class size is

(ii) The given classes are 10-19, 20-29, and 30-39. The classes can be written in continuous form as 9.5-19.5, 19.5-29.5, 29.5-39.5. The upper and lower limits of the first class are 19.5 and 9.5 respectively. Hence, the class size is

(iii) The given classes are 100-120, 120-140, 160-180, are in continuous form. The upper and lower limits of the first class are 120 and 100 respectively. Hence, the class size is

(iv) The given classes are 0-0.25, 0.25-0.50, 0.50-0.75, are in continuous form. The upper and lower limits of the first class are 0.25 and 0 respectively. Hence, the class size is

(v) The given classes are 5-5.01, 5.01-5.02, 5.02-5.03, are in continuous form. The upper and lower limits of the first class are 5.01 and 5 respectively. Hence, the class size is

Answer:

(i) The frequency distribution table of grouped frequency such as 30-39, 40-49… is the following:

(ii) The highest score is 100.

(iii) The lowest score is 37.

(iv) The range of the scores is

(v) If 40 is the pass mark, then the number of students failed is 2.

(vi) The number of students scored more than 75 is 8.

(vii) The observations 51, 54 and 57 in between 50-60 have not actually appeared.

(viii) Number of students scored less than 50 is 5.

Answer:

(i) The weights of the new born babies (in kg) in descending order are

3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.4, 2.4, 2.3, 2.2, 2.1

(ii) The highest weight is 3.1 kg.

(iii) The lowest weight is 2.1 kg.

(iv) The range of the weights is

(v) On that day, the number of babies born is 15.

(vi) Number of babies’ of weight less than 2.5 kg is 4.

(vii) Number of babies’ of weight more than 2.8 kg is 4.

(viii) Number of babies’ of weight 2.8 kg is 2.

Answer:

(i) After arranging in ascending order, the given data (runs) is 0, 0, 0, 15, 26, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 67, 71, 82, 84, 94, 105, 124 and 139.

(ii) The highest score of the player is 139.

(iii) The player did not score any run 3 times.

(iv) He scored 3 centuries.

(v) He scored more than 50 runs 12 times.

Answer:

Since, there are 7 class intervals and first class is 200–224, therefore all the class intervals are:

(i) 200–224

(ii) 225–249

(iii) 250–274

(iv)275–299

(v)300–324

(vi)325–349

(viii)350–374

We know that the class mark corresponding to each class mark is given by:

Now,

Therefore, the class marks are:

212, 237, 262, 287, 312, 337 and 362

Answer:

(i) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,

If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively

Lower limit of first class interval is;

And, upper limit of first class interval is:

Other class limits are:

(ii) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,

If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively

Lower limit of first class interval is;

And, upper limit of first class interval is:

Other class limits are:

(iii) Here, the data is distributed uniformly. So the class size h is given by difference between any two consecutive data. So,

If a is the class mark of a class interval and h be the class size, then the lower and upper limits of the class interval are: respectively

Lower limit of first class interval is;

And, upper limit of first class interval is:

Other class limits are:

Answer:

Here, the maximum and minimum values of the variate are 6 and 0 respectively.

So the range = 6 – 0 = 6

Here, we will take class size 1. So we must have 6 classes each of size 1.

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

Here, the maximum and minimum values of the variate are 95 and 29 respectively.

So the range = 95 – 29 = 66

Here, we will take class size 10. So we must have 7 classes each of size 10

Lower limit of first class interval is;

And, upper limit of first class interval is:

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

One of the class intervals is 160–164. This means that class size is 4

Here, the maximum and minimum values of the variate are 163 and 147 respectively.

So the range = 163 – 147 = 16

Here, we will take class size 4. So we must have 5 classes each of size 4

Lower limit of first class interval is;

And, upper limit of first class interval is:

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

Here, the maximum and minimum values of the variate are 898 and 804 respectively.

So the range = 898 – 804 = 94

Here, we will take class size 10. So we must have 94/10 i.e. 10 classes each of size 10.

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Lower limit of first class interval is;

And, upper limit of first class interval is:

Other class limits are:

Answer:

Here, the maximum and minimum values of the variate are 25.8 and 20.5 respectively.

So the range = 25.8 – 20.5 = 5.3

Here, we will take class size 1.

Lower limit of first class interval is;

And, upper limit of first class interval is:

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

Here, the maximum and minimum values of the variate are 320 and 210 respectively.

So the range = 320 – 210 = 110

Here, we will take class size 20 (As one class interval is 210–230).

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

It can be observed that 9 students have their blood group as A, 6 as B, 3 as AB, and 12 as O.

Therefore, the blood group of 30 students of the class can be represented as follows.

It can be observed clearly that the most common blood group and the rarest blood group among these students is O and AB respectively as 12 (maximum number of students) have their blood group as O, and 3 (minimum number of students) have their blood group as AB.

Answer:

By observing the data given above, the required frequency distribution table can be constructed as follows.

Answer:

(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..

The grouped frequency distribution table can be constructed as follows.

(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 − 20).

Answer:

Since the first class is –19.9 to –15

Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

Answer:

Cumulative frequency distribution is a table which displays the manner in which cumulative frequencies are distributed over various classes.

In a grouped frequency distribution the cumulative frequency of a class is the total of all frequencies upto and including that particular class. Make sure that, for calculating cumulative frequencies, the classes should be written in ascending order.

For example, the following table gives the cumulative frequency distribution of marks scored by 55 students in a test:

Cumulative frequency distributions are of two types, namely, less than and greater than or more than. For less than cumulative frequency distributions we add up the frequencies from the above and for greater than cumulative frequencies we add up the frequencies from below.

Answer:

Frequency distribution displays the frequencies of the corresponding class-intervals. But, the cumulative frequency distribution displays the cumulative frequencies of the corresponding classes.

For example, the following table gives the frequency and cumulative frequency distribution of marks scored by 55 students in a test:

In a frequency distribution, the sum of all the frequencies is equal to the total number of observations. But, in the cumulative frequency distribution, the last cumulative frequency is same as the total number of observations. For example, in the above table the sum of all the frequencies is 55, which is same as the total number of students and the last cumulative frequency is 55, which is same as the total number of students.

Answer:

The marks score by 55 students are given as

The cumulative frequency distribution is the following:

Answer:

The ages in years of 360 patients are given as

The cumulative frequency distribution is the following:

Answer:

The minimum and maximum bills are 24 Rs. and 95 Rs.

The range is.

Given that 70-79 is a class-interval. So, the class size is.

Now calculate

Thus, the number of classes is 8.

The cumulative frequency distribution is the following:

Answer:

The minimum and maximum numbers of books in shelves are 16 and 45.

The range is.

Given that 45-49 is the last class-interval. So, the class size is.

Now calculate

Thus, the number of classes is 8.

The cumulative frequency distribution is the following:

Answer:

We make class intervals 0-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55, 55-60, 60-65 and 65-70.

We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.

The frequency of the class 0-25 is 0.

The frequency of the class 25-30 is.

The frequency of the class 30-35 is.

The frequency of the class 35-40 is.

The frequency of the class 40-45 is.

The frequency of the class 45-50 is.

The frequency of the class 50-55 is.

The frequency of the class 55-60 is.

The frequency of the class 60-65 is.

The frequency of the class 65-70 is.

Here is the cumulative frequency distribution table.

Answer:

We make class intervals 0-240, 240-270, 270-300, 300-330, 330-360, 360-390 and 390-420.

(i) We will now find the frequencies of the different class-intervals.

Just proceed reverse to you proceed to generate cumulative frequency

Here is the frequency distribution table.

(ii) We are given the cumulative frequency table and we are asked to prepare new one.

So we will generate in some other fashion. Concentrate on frequency table and prepare the table. Here is the cumulative frequency distribution table.

Answer:

We make class intervals 0-12, 12-24, 24-36, 36-48, 48-60, 60-72, 72-84, 84-96 and 96-108.

We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.

The frequency of the class 0-12 is 1026−1124=98

The frequency of the class 12-24 is1026−809=217

The frequency of the class 24-36 is 809−427=382

The frequency of the class 36-48 is 427−158=269

The frequency of the class 48-60 is 158−20=138

The frequency of the class 60-72 is 20−5=15

The frequency of the class 72-84 is 5−3=2

The frequency of the class 84-96 is 3−1=2

The frequency of the class 96-108 is 1−0=1

Here is the cumulative frequency distribution table.

Answer:

Tally marks are used to find the frequencies.

Hence, the correct choice is (c).

Answer:

The difference between the highest and lowest values of the observations is called the range. Hence, the correct choice is (c).

Answer:

The difference between the upper and the lower class limits is called the class size. Hence, the correct choice is (b).

Answer:

The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30. Hence, the correct choice is (b).

Answer:

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is and the class-size is.

Therefore, we have two equations

Subtracting the second equation from the first equation, we have

Hence, the lower limit of the class is 13. Thus, the correct choice is (c).

Answer:

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is and the class-size is.

Given that the mid-value of the class is 42 and the class-size is 10. Therefore, we have two equations

Adding the above two equations, we have

Substituting the value of m in the first equation, we have

Hence, the upper and lower limits of the class are 47 and 37 respectively. Thus, the correct choice is (a).

Answer:

The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is (b).

Answer:

The number of classes is 9 and the uniform class size is 2.5. The lower limit of the lower class (first class) is 10.6. Therefore, the upper limit of the last class is

Hence, the correct choice is (b).

Answer:

The marks obtained by the students are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79 and 62.

The highest and lowest marks are 95 and 62 respectively. Therefore, the range of marks is

Hence, the correct option is (d).

Answer:

Tallies are usually marked in a bunch of 4. Hence, the correct option is (b). 

Answer:

Given that, the lower class limit of a class-interval is l and the mid-point of the class is m. Let u be the upper class limit of the class-interval. Therefore, we have

Thus the upper class limit of the class is. Hence, the correct choice is (c).

Answer:

The mid-value of a class is called the class mark.

The given class is 130–150.

∴ Class-mark of the given class $=\frac{130+150}{2}=\frac{280}{2}=140$

Thus, the class-mark of the class 130–150 is 140.

The class-mark of the class 130–150 is ___140___.

Answer:

Mid-value of the class = 10

Width of the class = 6

We know

Mid-value of the class = Lower limit of class + $\frac{1}{2}$(Width of the class)

∴ 10 = Lower limit of the class + $\frac{1}{2}$ × 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___.

Answer:

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__.

Answer:

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___.

Answer:

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 $a-\frac{h}{2}$ and $a+\frac{h}{2}$, respectively.

Here, a = 20 and h = 5.

∴ Lower class limit of the class having class mark 20 = $20-\frac{5}{2}=20-2.5=17.5$

Upper class limit of the class having class mark 20 = $20+\frac{5}{2}=20+2.5=22.5$

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__.

Answer:

If a is the mid-value of a class interval of width h, then the lower and upper limits of the class interval are $a-\frac{h}{2}$ and $a+\frac{h}{2}$, respectively.

Here, a = 42 and h = 10.

∴ Lower limit of the class interval = $42-\frac{10}{2}=42-5=37$

Upper limit of the class interval = $42+\frac{10}{2}=42+5=47$

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___.

Answer:

Mid-value of the class interval = m

Lower class limit of the class interval = l

We know

Mid-value of the class interval = $\frac{1}{2}$(Lower class limit + Upper class limit)

∴ m = $\frac{1}{2}$(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___.

Answer:

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,

$\frac{\mathrm{Range} \mathrm{of} \mathrm{the} \mathrm{data}}{\mathrm{Size} \mathrm{of} \mathrm{each} \mathrm{class} \mathrm{interval}}=\frac{98}{10}$ = 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___.

Answer:

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____.

Answer:

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____.

Answer:

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____.

Answer:

If a–b is a class in inclusive method, then in exclusive method the corresponding class is $\left(a-\frac{h}{2}\right)$–$\left(b+\frac{h}{2}\right)$, where h = $\frac{1}{2}$(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 = $\frac{1}{2}$ = 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, ...___.

Answer:

Mid-value of the class = m

Upper limit of the class = u

We know

Mid-value of the class = $\frac{1}{2}$(Lower limit of the class + Upper limit of the class)

∴ m = $\frac{1}{2}$(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____.