For a particular year, following is the distribution of ages (in years) of primary school teachers in a district: -Maths 9th

1 Answer

Answer :

1.First class interval is 15 – 20 and its lower limit is 15. 2.Fourth class interval is 30 – 35 Lower limit is 30 and upper limit is 35. 3.Class  mark  of the class 45 – 50 =( 45+50 )/ 2  =95 / 2  =47.5 4.Class size = Upper limit of each class interval – Lower limit of each class interval ∴ Here, class size = 20 – 15 = 5

Related questions

Description : For a particular year, following is the distribution of ages (in years) of primary school teachers in a district: -Maths 9th

Last Answer : 1.First class interval is 15 - 20 and its lower limit is 15. 2.Fourth class interval is 30 - 35 Lower limit is 30 and upper limit is 35. 3.Class mark of the class 45 - 50 =( 45+50 )/ 2 ... = Upper limit of each class interval - Lower limit of each class interval ∴ Here, class size = 20 - 15 = 5

Description : Five years hence , the age of Ram will be 10 more than the two thirds of Ravi’s age . Assume the present ages of Ram and Ravi as x and y respectively . Express the statement in the form of a linear equation in two variables. -Maths 9th

Last Answer : answer:

Description : A recent survey found that the ages of workers in a factory are distributed as follows: -Maths 9th

Last Answer : Total number of workers = 38 + 27 + 86 + 46 + 3 = 200 (i) P (person is 40 years or more) = P (Person having age 40 to 49 years) + P (person having age 50 to 59 years) + P (person having age 60 and ... 40 to 19 years) + P (person having age 50 to 59 years) = 86/200 + 46/200 = 132/200 = 0.66

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : Does your school district make publicly available the salaries of its teachers?

Last Answer : My district does not publish them, but you can go into the office and see them. By law they are a matter of public record in this state, so they have to be available for anyone to see.

Description : There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together. -Maths 9th

Last Answer : Total number of ways in which 10 person can sit around a circular table = 9! (∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 8 7 6 5 4 3 2 1) ways asthere is ... probability = \(rac{2 imes8!}{9!}\) = \(rac{2 imes8!}{9 imes8!}\) = \(rac{2}{9}.\)

Description : Find the number of ways in which 10 different flowers can be strung to form a garland so that three particular flowers are always together -Maths 9th

Last Answer : Consider the three particular flowers as one flower. Then we have (10 – 3) + 1 = 8 flowers which can be strung in the garland. Thus the garland can be formed in (8 – 1)!, i.e., 7! ways But the 3 particular flowers can be arranged amongst themselves in 3!

Description : In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time? -Maths 9th

Last Answer : ∴Required number of ways = 7C4 = 7!

Description : Eighteen quests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and -Maths 9th

Last Answer : Since four particular guests want to sit on particular side and three others on the other side. So, we are left with 11 guests out of which we choose 5 for side A in ways and remaining 6 for ... of the table can be arranged among themselves in 9! Ways. Hence, the total number of arrangements =

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Description : To draw a histogram to represent the following frequency distribution : -Maths 9th

Last Answer : Adjusted frequency of a class = Minimum class size of frequency distribution × Frequency of given class / Class size of given class ∴ Adjusted frequency for the class 25 - 45 = 5 × 8 / 20 = 2

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Description : Draw a histogram and frequency polygon for the following distribution : -Maths 9th

Last Answer : We represent class limits along x-axis and number of students along y-axis on a suitable Scale.

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Description : To draw a histogram to represent the following frequency distribution : -Maths 9th

Last Answer : Adjusted frequency of a class = Minimum class size of frequency distribution × Frequency of given class / Class size of given class ∴ Adjusted frequency for the class 25 - 45 = 5 × 8 / 20 = 2

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Description : Draw a histogram and frequency polygon for the following distribution : -Maths 9th

Last Answer : We represent class limits along x-axis and number of students along y-axis on a suitable Scale.

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Description : 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 -Maths 9th

Last Answer : NEED ANSWER

Description : 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. -Maths 9th

Last Answer : NEED ANSWER

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

Description : To draw a histogram to represent the following frequency distribution. -Maths 9th

Last Answer : NEED ANSWER

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : NEED ANSWER

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : NEED ANSWER

Description : 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 -Maths 9th

Last Answer : (b) Let x and y be the upper and lower class limit in a frequency distribution. Now, mid value of a class (x + y )/2=10 [given] ⇒ x + y = 20 (i) Also, given that, width of class x- y = 6 (ii) On ... putting x = 13 in Eq. (i), we get 13+y = 20 ⇒ y = 7 Hence, the lower limit of the class is 7.

Description : 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. -Maths 9th

Last Answer : (c) 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 ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : (b) Let x and y be the lower and upper class limit of a continuous frequency distribution. Now, mid-point of a class = (x + y)/2 = m [given] ⇒ x + y = 2 m =x + l = 2m [∴ y = l = upper class limit (given)] ⇒ x = 2 m-l Hence, the lower class limit of the class is 2m – l.

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

Description : To draw a histogram to represent the following frequency distribution. -Maths 9th

Last Answer : Frequency distribution.

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : No, here the widths of the rectangles are varying, so we need to make certain modifications in the length of the rectangles so that the areas are proportional to the frequencies. We proceed as ... Now, we get the following modified table So, the correct histogram with varying width is given below

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

Description : In a frequency distribution, -Maths 9th

Last Answer : Lower limit of the class = 10 - 1/2 x 6 = 10 - 3 = 7

Description : The frequency distribution: -Maths 9th

Last Answer : No, as the classes are of varying widths, not of uniform widths.

Description : The class marks of a continuous distribution are: -Maths 9th

Last Answer : It is not correct because the difference between two consecutive marks should be equal to the class size.

Description : If the class marks in frequency distribution are -Maths 9th

Last Answer : The class size of the distribution is = 40.5 - 33.5 = 7 The required class of the class mark 33.5 is [33.5 - 7/2] - [33.5 + 7/2], i.e., 30 - 37.