In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

1 Answer

Answer :

For Mean : As we know that  ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15   x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8  For Median : First of all arrange the data in the ascending order : 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98 Here, n= 15, an odd number  ∴  Median is the (n + 1 / 2)th observation or the (15 + 1 / 2)th observation or the 8th observation . ∴ Median = 52  For Mode :  Make a frequency table for given data :  Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52    

Related questions

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

Description : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25, 15, 20, 17, 10, 24, 15, 19 If a student is selected

Last Answer : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25 ... the probability that he gets more than 18 marks?

Description : The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th

Last Answer : In the given frequency distribution, the class intervals are not of equal width. Therefore, we would make modification in the lengths of the rectangle in the histogram so that the areas of rectangle ... draw rectangles with lengths as given in the last column. The histogram of data is given below:

Description : A problem in mathematics is given to four students A, B, C and D. Their chances of solving the problem -Maths 9th

Last Answer : Given, P(A) = \(rac{1}{2}\) ⇒ P(\(\bar{A}\)) = P(A not solving the problem) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)P(B) = \(rac{1}{3}\) ⇒ P(\(\bar{B}\)) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)P(C) = \ ... 3}{4}\)x \(rac{4}{5}\) = \(rac{1}{5}\)∴ P(problem will be solved) = 1 - \(rac{1}{5}\) = \(rac{4}{5}\).

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

Description : Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Last Answer : answer:

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

Description : The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Last Answer : Frequency Distribution Table Blood group O is most common as it has highest frequency, i.e., 12. Blood group AB is rarest as it has lowest frequency, i.e., 3.

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 : 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 : 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 : 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 diagnostic test in mathematics -Maths 9th

Last Answer : Median, because (i) each value occurs once (ii) the data is influenced by extreme values

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Description : The percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Last Answer : (i) Number of tests in which the student scored more than 70% marks = 3 ∴ P(more than 70% marks) = 3/6 = 1/2 (ii) Number of tests in which the student scored less than 70% marks = 3 ∴ P(less ... ) Number of tests in which the student scored at least 60% marks = 5 ∴ P(at least 60% marks) = 5/6

Description : Could you please upload video tutorials in Mathematics for our convinence -Maths 9th

Last Answer : NEED ANSWER

Description : Could you please upload video tutorials in Mathematics for our convinence -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Last Answer : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different. Hence, the total number of arrangements of ... 4!2!=12. Hence, the number of arrangement in which 4 vowels are together =(10080×12)=120960.

Description : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Last Answer : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, ... , 23, 35. Find the average mark of the students.

Description : The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below :   The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15 

Last Answer : Answer: A

Description : A class has `175` students. The following data shows the number of students obtaining one or more subjects : Mathematics `100`, Physics `70`, Chemistr

Last Answer : A class has `175` students. The following data shows the number of students obtaining one or ... . How many students have offered Mathematics alone ?

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

Description : Two coins are tossed 1000 times and the outcomes are recorded as below: -Maths 9th

Last Answer : P (at most one head) = P (0 head) + P (1 head) = 250/1000 + 550/1000 = 800/1000 = 4/5

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

Description : You have used a spreadsheet to calculate the average marks of a class in Mathematics. -Technology

Last Answer : (b) Marks of each student tor Mathematics will be required for calculating average marks of mathematics.

Description : Nandita scores 60% marks in five subjects together, viz., Hindi, Science, Mathematics, English and Sanskrit, where in the maximum marks of each subject were 105. How many marks did Nandita score in Science, if she scored 69 ... and 51 marks in English? a) 66 b) 68 c) 55 d) 65 e) None of these

Last Answer : Answer: D Total of maximum marks of all subjects=105×5=525 75% of 525=525 ×60/100= 315 Obtained marks of foru subjects (Hindi, Sanskrit, mathematics and English) =69+62+68+51=250 So, the obtained marks in Science=315-250=65

Description : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

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 : how i can score more marks in sa1 -Maths 9th

Last Answer : u need to focus on ncert book than iinterfaces onlin if ur weak in thes u should do these math example science pratical social explanation surf internet on ur topic hindi open up with others inspeeaking english same

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 : 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 : 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.