Скачать или смотреть EX 14.2, COMPLETE EXERCISE IN ONE VIDEO, CH.14, 10TH CLASS, NCERT, CBSE, STATISTICS सांख्यिकी ❤️❤️🔥🔥

CLASS 10TH
SUBJECT MATH
IN BOTH MEDIUM
CHAPTER 14
EXERCISE 14.2
COMPLETE SOLUTIONS OF EXERCISE

EX 14.1, COMPLETE EXERCISE IN ONE VIDEO, CH.14, 10TH CLASS, NCERT, CBSE, STATISTICS सांख्यिकी ❤️❤️🔥🔥
   • EX 14.1, COMPLETE EXERCISE IN ONE VIDEO, C...  

EX 14.2, COMPLETE EXERCISE IN ONE VIDEO, CH.14, 10TH CLASS, NCERT, CBSE, STATISTICS सांख्यिकी ❤️❤️🔥🔥
   • EX 14.2, COMPLETE EXERCISE IN ONE VIDEO, C...  


EX 14.3, COMPLETE EXERCISE IN ONE VIDEO, CH.14, 10TH CLASS, NCERT, CBSE, STATISTICS सांख्यिकी ❤️❤️🔥🔥
   • EX 14.3, COMPLETE EXERCISE IN ONE VIDEO, C...  


14.1Introduction In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions. You have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths) and frequency polygons. In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data. We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives. 14.2Mean of Grouped Data The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if x1 , x2 ,. . ., xn are observations with respective froccurs f2 times, and so on. Now, the sum of the values of all the observations = f1 x1 + f2 x2 + . . . + fn xn , and the number of observations = f1 + f2 + . . . + fn . So, the mean x of the data is given by x =1122 12 nn n fxfxfx fff +++ +++   Recall that we can write this in short form by using the Greek letter Σ (capital sigma) which means summation. That is,
  which, more briefly, is written as x = Σ Σii i fx f, if it is understood that i varies from 1 to n. Let us apply this formula to find the mean in the following example. Example 1 : The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. Marks obtained10203640505660707280889295 (xi ) Number of1134324411231 students (fi ) Solution: Recall that to find the mean marks, we require the product of each xi with the corresponding frequency fi . So, let us put them in a column as shown in Table 14.1. Table 14.1 Marks obtained (xi ), the mean marks obtained is 59.3. In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean. Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table (see Table 14.2). Table 14.2 Class interval10 - 2525 - 4040 - 5555 - 7070 - 8585 - 100 Number of students237666 Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is, Class mark =Upperclasslimit+Lowerclasslimit 2 With reference to Table 14.2, for the class 10-25, the class mark is 1025 2 +, i.e., 17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table 14.3. These class marks serve as our xi ’s. Now, in general, for the ith class interval, we have the frequency fi corresponding to the class mark xi . We can now proceed to compute the mean in the same manner as in Example 1.
The sum of the values in the last column gives us Σ fi xi . So, the mean x of the given data is given by x =1860.06230ii i fx f Σ==Σ This new method of finding the mean is known as the Direct Method. We observe that Tables 14.1 and 14.3 are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 14.3, 59.3 being the exact mean, while 62 an approximate mean. Sometimes when the numerical values of xi and fi are large,