ID : ph-9-data-handling-probability-statistics [1] Grade 9 Data Handling - Probability, Statistics For more such worksheets visit www.edugain.com Answer t he quest ions (1) Danilo selects 3 numbers randomly f rom the f ollowing set of 5 numbers 1, 7, 9, 8 and 2. He puts them in the f orm of a proper f raction of the type a b c. What is the probability that you will get a f raction greater than 2 21 22? (2) From among a group of 4 men, 3 women and 4 children, 2 individuals are selected randomly. What is the probability that exactly 1 among the chosen are children? (3) If there are n numbers of which one is and all the others are 1's, then by how much is the arithmetic mean of these numbers less than 1. (4) Perla got an average score of 80.5 in 4 tests. She got 81 as the average of the highest 3 scores, and his lowest two scores are the same numbers. What is the average of her highest two scores? Choose correct answer(s) f rom given choice (5) Find the median of the f ollowing set 10, 22, 10, 21, 13, 20, 15, 13 a. 13 b. 12 c. 15 d. 14 (6) What is the average of the 11 consecutive integers starting at 15? a. 20 b. 19 c. 17 d. 21 (7) If 80C - 240A = 518, and 320A + 80B = 922, then what is the average of A, B and C? a. 6 b. 8 c. 4 d. 7 (8) If a prime number is less than 31, what is the probability that it is also less than 23. a. 7/10 b. 4/5 c. 8/11 d. 9/10
ID : ph-9-data-handling-probability-statistics [2] (9) A poll is taken among 10000 people working in Antipolo. The aim was to see what their annual salaries were. Annual Salary Number of people Less than 40000 180 40001 to 75000 740 75001 to 150000 980 150001 to 250000 3240 More than 250000 4860 If you choose a person at random f rom this group, what is the probability that he or she earns more than 75000 annually? a. 9097 10000 b. 9080 10000 c. 9044 10000 d. 9043 10000 (10) In 2008, the meteorological of f ice predicted the weathers completely right f or the months of f ebruary and april, and completely wrong f or all the other months. What is the probability that the f orecast was wrong f or a given day that year? a. 308 366 c. 307 365 b. 305 365 d. 307 366 (11) Coin A is f lipped 3 times and coin B is f lipped 5 times. What is the probability that the number of heads obtained f rom f lipping the two coins is the same? a. 3 30 b. 9 32 c. 9 34 d. 7 32 (12) Perla's age is twice that of Lualhati, Lualhati's age is twice that of Danilo and Danilo's age is twice that of Arvin. If the average of the ages of Perla, Lualhati, Danilo and Arvin is 30 years, then what is the present age of Lualhati. a. 32 years b. 33 years c. 31 years d. 30 years
ID : ph-9-data-handling-probability-statistics [3] Fill in the blanks (13) The f ollowing are the marks obtained by 40 students in Geography. Marks T ally Marks Number of students 2 3 3 8 4 5 5 5 6 5 7 5 8 7 9 2 The median of their score is (14) The average of any f our consecutive odd integers is always. (15) In an of f ice, the age of the employees was as f ollows 28, 39, 51, 25, 31, 40, 28, 33, 52, 27, 53 The average age of the employees is 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ph-9-data-handling-probability-statistics [4] (1) 18 30 We need to select 3 numbers out of the 5 given in order to get a f raction of the f orm a b c We are also told it is a proper f raction, so b should be greater than c Let's put the integers in a sorted manner. We get 1, 2, 7, 8 and 9 Now we need to see how many proper f ractions can be f ormed f rom them A proper f raction of the type a b c has three integers, a whole number a, a numerator b, and a denominator c Let's assume we use one of the integers in the list above as the whole number a We now can select b and c f rom the remaining 4 integers We can select 2 integers f rom 4 in 4C2 = 4x3 = 6 ways 2x1 For any pair we select, one will be greater than the other, and the smaller integer will f orm the numerator and the larger one the denominator - Note: that the other way won't work - if the numerator is larger than the denominator it is not a proper f raction So f or each of the 5 integers, if we select one as the whole number, we get 6 possible combinations of numerator and denominator that can f orm a proper f raction This means there are 5x6 = 30 possible proper f ractions of the f orm a b c that can be f ormed f rom these 5 integers Step 4 Now we need to f igure out how many of these 30 f ractions are greater than 2 21 22 In gt, we see that the whole number is 2, the numerator is 21 and the denominator is 22 We see that the numerator 21 is larger than the largest number in the set of numbers given to us, and the denominator 22 is one larger than 21 The implication of this is that 2 21 will be larger than any f raction that can be f ormed 22 f rom the set of numbers 1, 2, 7, 8 and 9 where the whole number of the f raction is 2 So we only need to count the f ractions that have the whole number greater than 2 These are the f ractions that will have the whole numbers as 7, 8 and 9 Step 5
ID : ph-9-data-handling-probability-statistics [5] Now, remember there are 30 proper f ractions you can f orm f rom this list of number From our analysis above, we also saw that f or each number selected as the whole number, we can f orm 6 f ractions f rom this list of numbers Step 6 So using each of the numbers f rom 7, 8 and 9 as the whole number, we can f orm 6 proper f ractions The total number of f ractions that can be f ormed using 7, 8 and 9 as the whole number = 6 x 3 = 18 Step 7 Out of the 30 f ractions, 18 will be greater than 2 21 22 Step 8 The probability is theref ore = 18 30
(3) 1 ID : ph-9-data-handling-probability-statistics [6] n 2 It is given that there are n numbers of which one is and all the others are 1's. Theref ore, the numbers are, 1, 1, 1 (where n is the total number of numbers in the series) Out of n numbers one is and remaining n-1 numbers are 1. Theref ore, the sum of n-1 numbers is n-1. Now, the sum of all numbers in the series = n-1 + (1-1 n ) = n - 1 n Now, the arithmetic mean of the numbers = Sum of the all numbers n = n - 1 n n = n n = 1 - - 1 n 2 1 n 2 Step 4 Thus, we can say that the arithmetic mean of these numbers is 1 n 2 less than 1. (4) 82
(5) d. 14 ID : ph-9-data-handling-probability-statistics [7] If you look at the question caref ully, you will notice that the given data is 10, 22, 10, 21, 13, 20, 15, 13 To f ind the median f irst of all arrange the data in the ascending order, you get 10, 10, 13, 13, 15, 20, 21, 22 Total number of terms are 8 which is even. So, median is equal to the average of ( n 2 ) th and ( n 2 +1) th terms, (where n is the number of terms) ( n 2 ) th = ( 8 2 ) th = 4 th ( n 2 +1) th = ( 8 2 + 1) th = 5 th 4 th and 5 th terms are 13 and 15 respectively 13 + 15 median = 2 = 28 2 = 14 Step 4 Theref ore the median of the data set is 14.
(6) a. 20 ID : ph-9-data-handling-probability-statistics [8] If you look at the question caref ully, you will notice that the 11 consecutive integers starting at 15 = 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 Sum of 11 consecutive integers = 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 = 220 Average of 11 consecutive integers starting at 15 = Sum of 11 consecutive integers T otal number of consecutive integers = 220 11 = 20 Now the average of 11 consecutive integers at 15 is 20. (7) a. 6 According to the question, 80C - 240A = 518 -----(1), 320A + 80B = 922 -----(2) By adding equation (1) and (2), we get: 80C + 320A -240A + 80B = 1440 80C + 80A + 80B = 1440 80(C + A + B) = 1440 C + A + B = 1440/80 A + B + C = 18 Now, the average of A, B, C = A + B + C 3 = 18 3 = 6
(8) b. 4/5 ID : ph-9-data-handling-probability-statistics [9] There are 10 prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23, 29) which are less than 31 Of these 8 are less than 23 Theref ore probability is 8/10 = 4/5 (9) b. 9080 10000 First we need to f ind the total number of people among whom the poll was taken. We add the number of people in the various sets 180 + 740 + 980 + 3240 + 4860 = 10000. To f ind the probability that the the random person chosen has a salary of more than 75000, we need to add the number of people who have salaries greater than this number. This is 980 + 3240 + 4860 = 9080. Step 4 So, 9080 people out of a total of 10000 earn an annual salary greater than 75000. The probability that the randomly chosen person has a salary greater than 75000 = 9080. 10000
(10) d. 307 366 ID : ph-9-data-handling-probability-statistics [10] The key thing to note here is that 2008 is a leap year A leap year has 366 days Now we need to f igure out the number of days in f ebruary and april f ebruary has 29 days and april has 30 days Adding them together we get 29 + 30 = 59 days Step 4 So the f orecast was right f or 59 days and wrong f or 307 days Step 5 The probability that the f orecast was wrong on a given day would theref ore be 307 366 (11) d. 7 32 (12) a. 32 years
(13) 5 ID : ph-9-data-handling-probability-statistics [11] Marks T ally Marks Number of students 2 3 3 8 4 5 5 5 6 5 7 5 8 7 9 2 If you look at the given table f rom top to bottom caref ully, you will notice that the marks obtained by 40 students in Geography are arranged in ascending order. Total number of students are 40 which is even. So, median is equal to the average of ( n 2 ) th and ( n 2 +1) th student's marks in Geography, (where n is the number of students) ( n ) th = ( 40 ) th = 20 th 2 2 ( n +1) th = ( 40 + 1) th = 21 th 2 2 If you count the number of students in tally marks column of the table, you will notice that the marks obtained by 20 th and 21 th students in Geography are 5 and 5 respectively median = 5 + 5 2 = 10 2 = 5 Now the median of their scores is 5.
(14) even ID : ph-9-data-handling-probability-statistics [12] The average of any f our consecutive odd integers is always even. For example 1, 3, 5 and 7 are the f our consecutive odd integers, average of 1, 3, 5 and 7 = 1 + 3 + 5 + 7 4 = 16 4 = 4 Which is an even number. Theref ore we can say that the average of any f our consecutive odd integers is always an even. (15) 37 If you look at the question caref ully, you will notice that the age of the employees was as f ollows 28, 39, 51, 25, 31, 40, 28, 33, 52, 27, 53 Sum of age of the employees = 28 + 39 + 51 + 25 + 31 + 40 + 28 + 33 + 52 + 27 + 53 = 407 Total number of employees = 11 Average age of the employees = Sum of age of the employees T otal number of employees = 407 11 = 37 Now the average age of the employees is 37.