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Statistics
1.0Introduction
It is the branch of mathematics which deals with the collection, presentation, analysis and interpretation of numerical data.
In singular form, statistics is taken as a subject. And, in plural form, statistics means data. Class interval: Each group into which the raw data is condensed, is called a class-interval. Class limits: Each class is bounded by two figures, which are called class limits. The figure on the left side of a class is called its lower limit and that on its right is called its upper limit.
Exclusive form (or continuous interval form): A frequency distribution in which the upper limit of each class is excluded and lower limit is included, is called an exclusive form.
Example: Suppose the marks obtained by some students in an examination are given. We may consider the classes etc. In class , we include 0 and exclude 10. In class , we include 10 and exclude 20.
Inclusive form (or discontinuous interval form): A frequency distribution in which each upper limit as well as lower limit is included, is called an inclusive form. Thus, we have classes of the form etc.
In , both 0 and 10 are included.
2.0Important terms related to grouped data:
Class boundaries or true upper and true lower limits: (i) In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit. (ii) In the inclusive form, the number midway between the upper limit of a class and lower limit of the subsequent class gives the true upper limit of the class and the true lower limit of the subsequent class. Thus, in the above table of inclusive form, we have: true upper limit of class is and true lower limit of class is 10.5 . Similarly, true upper limit of class is , and true lower limit of class is 20.5 .
Class size The difference between the true upper limit and the true lower limit of a class is called its class size.
Class mark of a class Class mark
- The difference between any two successive class marks gives the class size.
- Average is the statistic which describes the center of a set of data, a set of numbers which are measurements or counts.
The most commonly used averages are the mean (arithmetic average), mode (most frequent number), median (middle number when numbers are listed smallest to largest).
Numerical Ability 1 The class marks of a frequency distribution are . Find the class-size and all the class-intervals. Solution: Class size Difference between two successive class-marks . Let the lower limit of the first-class interval be . Then, its upper limit . So, the first class-interval is 4-10. Let the lower limit of last class-interval be . Then, its upper-class limit . . So, the last class-interval is 40-46. Hence, the required class-intervals are and . An average tends to lie centrally with the values of the variable arranged in ascending order of magnitude. So, we call an average a measure of central tendency of the data. Three measures of central tendency are useful for analysing the data, namely (a) Mean (b) Median (c) Mode
- We know that the mean of observations is the sum of the values of all the observations divided by the total number of observations i.e., if are n observations, then mean or , where denotes the sum .
3.0Arithmetic mean
The arithmetic mean of grouped data may also be calculated by any one of the following methods:
- Direct method
- Assumed-mean method
Direct method
are observations with respective frequencies then mean, is defined by or , Where The following steps should be followed in finding the arithmetic mean of grouped data by direct method. Step-1: Find the class mark ( ) of each class using, Step - 2: Calculate for each i Step - 3: Use the formula : mean ,
- Sum of first ' ' natural numbers is
Numerical Ability 2 Find the mean of the following data:
Solution: We may prepare the table as given below:
Mean,
Numerical Ability 3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency .
Solution: We may prepare the table as given below:
Given, mean
- The sum of squares of ' ' natural numbers is
Numerical Ability 4 Find the missing frequencies and in the table given below, it is being given that the mean of the given frequency distribution is 50 .
Solution: We may prepare the table as given below:
Mean, Given, mean And Adding (i) and (ii), we get Hence, the missing frequencies and are 28 and 24 respectively.
Numerical Ability 5 Find the mean of the following distribution.
Solution:
Mean,
Numerical Ability 6 Find the mean marks from the following data:
Solution: We may prepare the table as given below:
Mean,
If less than type or more than type frequency distribution table is given then convert table is usual form to find the mean.
Numerical Ability 7 Find the mean marks of students from the adjoining frequency distribution table.
Solution: We may prepare the table as given below:
Mean, (approx)
Assumed mean method
In this case, to calculate the mean, we follow the following steps: Step - 1: Find the class mark of each class using Step - 2 : Choose a suitable value of in the middle as the assumed mean and denote it by 'a'. Step-3: Find for each i Step-4: Find for each i Step-5: Find Step - 6 : Calculate the mean, by using the formula
Numerical Ability 8 The following table gives the marks scored by 100 students in a class test:
Find the mean marks scored by a student in class test.
Solution: We may prepare the table with assumed mean, as given below:
Mean,
Numerical Ability 9 Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, by using assumed-mean method.
Solution: We may prepare the table with assumed mean, as given below:
Mean,
Numerical Ability 10 Find the arithmetic mean of the following frequency distribution.
Solution: The given series is in inclusive form. We will prepare the table in exclusive form with assumed mean a as given below:
- It is not necessary to convert the data in continuous interval form for finding mean.
Step-deviation method or short-cut method
Sometimes, the values of x and f are so large that the calculation of mean by assumed mean method becomes quite inconvenient. In this case, we follow the following steps: Step-1 : Find the class mark of each class by using Step-2 : Choose a suitable values of in the middle as the assumed mean and denote it by 'a'. Step-3 : Find (upper limit -lower limit) for each class. Step-4 : Find for each class. Step-5 : Find for each i. Step-6 : Calculate, the mean by using the formula , where .
Numerical Ability 11 Find the mean of the following distribution by step-deviation method:
Solution: We may prepare the table with assumed mean and as given below:
- If class size is different then h is taken as HCF of the class marks.
Numerical Ability 12 Calculate the mean for the following frequency distribution (By step deviation method).
Solution:
Mean
4.0Median
Median: It is a measure of central tendency which gives the value of the middle most observation in the data. In a grouped data, it is not possible to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves.
Median Class: The class whose cumulative frequency is greater than is called the median class.
To calculate the median of a grouped data, we follow the following steps.
Step - 1: Prepare the cumulative frequency table corresponding to the given frequency distribution and obtain
Step-2: Find
Step - 3: Look at the cumulative frequency just greater than and find the corresponding class (Median class).
Step - 4: Use the formula Median Where, Lower limit of median class. Frequency of the median class. Cumulative frequency of the class preceding the median class. Size of the median class. .
- Data must be in continuous interval form to find median and mode of grouped data.
Numerical Ability 13 Find the median of the following frequency distribution:
Solution: At first, we prepare a cumulative frequency distribution table as given below:
Here, N = 100 The cumulative frequency just greater than 50 is 64 and the corresponding class is . So, the median class is .
- The median is the middle of a distribution: half the scores are above the median and half are below the median.
Numerical Ability 14 A health insurance agent found the following data for distribution of ages of policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
Solution: From the given table we can find the frequency and cumulative frequencies as given below :
Here, The cumulative frequency just greater than 50 is 78 and the corresponding class is . So, the median class is . Therefore, median Hence, the median age is 35.76 years.
Numerical Ability 15 From the following frequency distribution, calculate the median.
Solution: Here The median lies in the class . , c.f. and
Calculation of Median
Applying the formula, Median We get, Median So, above half the student have scored marks less than 27.74 and the other half scored marks more than 27.74.
Numerical Ability 16 Calculate the missing frequency 'a' from the following distribution, it is being given that the median of the distribution is 24 .
Solution: At first we prepare a cumulative frequency distribution table as given below :
Since the median is 24, therefore, the median class will be . Hence, and Therefore, median Hence, the value of missing frequency a is 25 .
Numerical Ability 17 The median of the following data is 525 . Find the values of and , if the total frequency is .
Solution: At first we prepare a cumulative frequency distribution table as given below :
We have Since the median is 525 , so, the median class is 500-600 and Therefore, median Also, putting in (i), we get Hence, the values of and are 9 and 15 respectively.
5.0Mode
Mode : Mode is that value among the observations which occurs most often i.e. the value of the observation having the maximum frequency.
In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies.
Modal Class : The class of a frequency distribution having maximum frequency is called modal class of a frequency distribution.
The mode is a value inside the modal class and is calculated by using the formula. Mode Where Lower limit of the modal class. h = Size of class interval Frequency of modal class Frequency of the class preceding the modal class Frequency of the class succeeding the modal class.
- A disadvantage of the mode is that many distributions have more than one mode. These distribution are called "multi modal" and is therefore not recommended to be used as the only measure of central tendency.
Numerical Ability 18 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Determine the modal lifetime of the components.
Solution: Here the class 60-80 has maximum frequency, so it is the modal class. and Therefore, mode Hence, the modal lifetime of the components is 65.625 hours.
Numerical Ability 19 Given below is the frequency distribution of the heights of players in a school.
Find the modal class.
Solution: The given series is in inclusive form. We prepare the table in exclusive form, as given below :
Here, the class 165.5-168.5 has maximum frequency, so it is the modal class.
Numerical Ability 20 The mode of the following series is 36 . Find the missing frequency in .
Solution: Since the mode is 36 , so the modal class will be and Therefore, mode Hence, the value of the missing frequency is 10 .
Numerical Ability 21 The mean and mode of a frequency distribution are 28 and 19 respectively. Then find the median. Solution: Mean and Mode Median ? Mode Median -2 Mean Median - 2(28) 3 Median = Median Median
6.0Graphical representation of cumulative frequency distribution
Cumulative frequency polygon curve (Ogive) Cumulative frequency is of two types and corresponding to these, the ogive is also of two types. • Less than ogive • More than ogive
Less than ogive: To construct a cumulative frequency polygon and an ogive, we follow these steps : Step-1: Mark the upper class limit along x -axis and the corresponding cumulative frequencies along -axis. Step-2: Plot these points successively by line segments. We get a polygon, called cumulative frequency polygon. Step-3: Plot these points successively by smooth curves, we get a curve called cumulative frequency curve or an ogive.
More than ogive: To construct a cumulative frequency polygon and an ogive, we follow these steps: Step-1: Mark the lower class limits along -axis and the corresponding cumulative frequencies along -axis. Step-2: Plot these points successively by line segments, we get a polygon, called cumulative frequency polygon. Step-3: Plot these points successively by smooth curves, we get a curve, called cumulative frequency curve or an ogive.
Application of an ogive
Ogive can be used to find the median of a frequency distribution. To find the median, we follow these steps.
Method-I
Step-1: Draw anyone of the two types of frequency curves on the graph paper. Step-2: Compute and mark the corresponding points on the -axis. Step-3: Draw a line parallel to -axis from the point marked in step 2, cutting the cumulative frequency curve at a point . Step-4: Draw perpendicular PM from on the -axis. The x -coordinate of point M gives the median.
Method-II
Step-1: Draw less than type and more than type cumulative frequency curves on the graph paper. Step-2: Mark the point of intersecting (P) of the two curves drawn in step 1. Step-3: Draw perpendicular PM from on the -axis. The - coordinate of point gives the median.
Numerical Ability 22 The following distribution gives the daily income of 50 workers of a factory.
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Solution: From the given table, we prepare a less than type cumulative frequency distribution table, as given below:
Now, plot the points and . Join these points by a freehand curve to get an ogive of 'less than' type.
Numerical Ability 23 The following table gives the weight of 120 articles :
Change the distribution to a 'more than type' distribution and draw its ogive.
Solution:
Plotting the points :
Numerical Ability 24 The annual profits earned by 30 shops of a shopping complex in a locality gives rise to the following distribution:
Draw both ogives for the data above. Hence, obtain the median profit.
Solution: We have a more than type cumulative frequency distribution table. We may also prepare a less than type cumulative frequency distribution table from the given data, as given below:
Now, plot the points A , B , C and for the more than type cumulative frequency and the points and for the less than type cumulative frequency distribution table. Join these points by a freehand to get ogives for 'more than' type and 'less than' type.
Numerical Ability 25 The following data gives the information on marks of 70 students in a periodical test: Draw a cumulative frequency curve for the given data and find the median.
Solution: We have a less than cumulative frequency table. We mark the upper class limits along the x -axis and the corresponding cumulative frequencies (no. of students) along the y -axis. Now, plot the points and . Join these points by a freehand curve to get an ogive of 'less than' type.
7.0Memory map