Grade 9 Data Handling - Probability, Statistics

Transcription

1 ID : ae-9-data-handling-probability-statistics [1] Grade 9 Data Handling - Probability, Statistics For more such worksheets visit Answer t he quest ions (1) The numbers 1 to 19 are written on 19 pieces of paper and dropped into a box. Three of them are drawn at random. What is the probability that the three pieces of paper picked have numbers that are in arithmetic progression? () The f ollowing table shows prof its of 5 companies in a business group last year. Company Profit Company 1 Dhs 500,000 Company Dhs 00,000 Company 3 Dhs 400,000 Company 4 Dhs 00,000 Company 5 Dhs 50,000 What is the average prof it of the business group? (3) There are 3 badges belonging to 3 students. The badges were put into a box, and each student pulls out an badge one af ter the other. What is the probability that each student gets his or her own badge? (4) Find the median of the f ollowing set 1, 14, 11, 1, 13, 1, 0, 0, 14, 13, 13 (5) Two dice are rolled. What is the probability that the two numbers add up to a prime number? Choose correct answer(s) f rom given choice (6) Kadin is part of the school cricket team, and this year he has scored an average of 3 runs. He has played 5 innings so f ar, and his scores in 4 of them are 6, 38, 30, 7 What was his score in the last one? a. 39 b. 36 c. 40 d. 33

2 (7) What is the probability that an integer in the set 1,, 3, is divisible by and not divisible by 3? ID : ae-9-data-handling-probability-statistics [] a b c d (8) 6 coins are tossed in parallel. What is the probability of getting at least one tails? a b c d (9) In an of f ice, the age of the employees was as f ollows 41, 36, 47, 3, 48, 41, 7, 31, 6, 35, 43 What is the average of their ages? a. 3 b. 37 c. 40 d. 34 (10) In a class of 0 students, the average marks in a test was 71. If Givon got 7 marks and Kamal got 106 marks, what was the average of the other 18 students? a. 7 b. 68 c. 69 d. 65 (11) A tyre manuf acturer keeps the record of how much distance the tyres manuf actured by the company travel bef ore f ailing. They f ind the f ollowing data Distance traveled in kilometers Number of failing tyres Less than to to to More than If Anan buys a tyre f rom them, what is the probability that it will last more than 0000 kilometers? a c b d

3 (1) What is the probability that a leap year will contain 53 Sundays? ID : ae-9-data-handling-probability-statistics [3] a. 366 b. 1 5 c. 1 7 d. 7 Fill in the blanks (13) The average of any f our consecutive odd integers is always. (14) Four children are playing in a park. Aziz is 1 years and 4 months old. Kalil is 1 years and 5 months old. Ghassan is years old, and Hadya is 1 1 years old. Their average age is years and months. (15) The f ollowing are the marks obtained by 30 students in German. Marks T ally Marks Number of students The median of their score is 016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : ae-9-data-handling-probability-statistics [4] (1) 7 33 This is a little complicated, so f ollow caref ully For making the explanation and the equations simpler, think of the number on the pieces of paper in the f orm of (n+1) Here, we can see f rom the equation n+1 = 19, so n=9 The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between 1 and 19 is the ratio of - Number of ways we can get an A.P f rom 3 random numbers between 1 to 19, and - Number of ways to select 3 random numbers between 1 to 19 Step 4 Let's look at the second part f irst. Three tickets can be drawn f rom (n+1) numbers is in [( x n) + 1] C 3 ways i.e. Number of ways 3 tickets can be drawn = (n+1)(n)(n-1) 3xx1 Simplif ying this, we get the number of ways to draw 3 numbers between 1 and 19 = n(4n -1) 3 Here n = 9, so we can simplif y it as 969 Step 5 Now f or the ways we can get an A.P f rom 3 numbers bwetween Arithmetic Progressions of 3 numbers would be a sequence of 3 numbers that are separated by a common interval e.g. 1,,3 or 3,5,7 etc. They are in the f orm (a, a+d, a + d), where a is an integer f rom 1 to (19-), and d is another integer So it's helpf ul to think of the solution in terms of this interval. So we'll think of all the sequences that have an interval 1, then sequences with interval, and so on Step 6 So what are the possible sequences with interval 1. They are (1,,3) (,3,4)... (n-1,n,n+1) T here are theref ore n-1 such possible sequences Step 7 Similarly, let's look at A.P with interval between the terms. They are

5 (1,3,5) (,4,6)...<n-3,n-1,n+1> T here are n-3 such possible sequences ID : ae-9-data-handling-probability-statistics [5] Step 8 We can generalize this to say that the number of such sequences with interval 'd' is (n- (d-1)) Obviously the largest possible integer is d=n, with just one sequence (1, n+1, n+1) Step 9 So the total number of such sequences is (n-1) + (n-3) + (n-5) This is itself an AP with n terms and d= The sum of this sequence is n [ + (n-1)] Simplif ying, we get n Here n=9, so this is 81 0 So the probability is 7 33

6 () Dhs ID : ae-9-data-handling-probability-statistics [6] If you look at the question caref ully, you will notice that the table below shows prof its of 5 companies in a business group last year. Company Profit Company 1 Dhs 500,000 Company Dhs 00,000 Company 3 Dhs 400,000 Company 4 Dhs 00,000 Company 5 Dhs 50,000 Total prof its made by the business group = Dhs 500,000 + Dhs 00,000 + Dhs 400,000 + Dhs 00,000 + Dhs 50,000 = Dhs Average prof it of the business group = Total prof its made by the business group Number of companies in a group = = Now the average prof it of the business group is Dhs (3) 1 6 We have 3 badges and 3 students. Let's f irst see in how many dif f erent ways can the badges be distributed among the students We know that 3 objects can be distributed in 3! = 3 x x...x 1 = 6 ways Out of these 6 ways, there is only one distribution where each student got his or her own badge Step 4 So the probability of each student getting his or her own badge = 1 6

7 (4) 14 ID : ae-9-data-handling-probability-statistics [7] If you look at the question caref ully, you will notice that the given data is 1, 14, 11, 1, 13, 1, 0, 0, 14, 13, 13 To f ind the median f irst of all arrange the data in the ascending order, you get 11, 1, 13, 13, 13, 14, 14, 0, 0, 1, 1 Median is the middle number in a sorted list. Total number of terms are 11 which is odd. (n + 1) So, Median = ( ) th (where n is the number of terms) =( ) th = ( 1 ) th = 6 th Since 6 th term in the data 11, 1, 13, 13, 13, 14, 14, 0, 0, 1, 1 is 14, theref ore Median = 14 Step 4 Theref ore the median of the data set is 14.

8 (5) 15 ID : ae-9-data-handling-probability-statistics [8] 36 The two dice that are rolled can show any of these values Dice 1 : 1,, 3, 4, 5, 6 Dice : 1,, 3, 4, 5, 6 So we can get a total of 36 combinations between them (6 x 6) If we take one value f rom the list of possible values f rom each Dice, we get numbers ranging f rom (when both Dice show 1) to 1 (when both dice show 6). Let's enumerate the prime numbers between and 1. They are, 3, 5, 7 and 11 We need to see in how many ways we can get each of these values Let's put the value rolled by the dice as (x,y), where x is the value rolled by Dice 1, and y the value rolled by Dice - : The only way to get this is when we roll (1,1). 1 possibility - 3: We can get this by (1,) or (,1). possibilities. - 5: We can get this by (,3), (3,), (1,4) or (4,1). 4 possibilities. - 7: We can get this by (1,6), (,5), (3,4), (4,3), (5,), (6,1). 6 possibilities. - 11: We can get this by (5,6), or (6,5). Possibilities This gives us a total of = 15 possible ways to get a prime number So the probability of getting the two numbers add up to a prime is 15 36

9 (6) a. 39 ID : ae-9-data-handling-probability-statistics [9] If you look at the question caref ully, you will notice that Kadin has scored an average of 3 runs in 5 innings. Let Kadin score in the last inning = x His scores in 4 innings out of 5 innings = 6, 38, 30, 7 Average score = Total score in 5 innings total innings 3 = x 5 3 = 11 + x = 11 + x 160 = 11 + x 11 + x = 160 x = x = 39 Theref ore Kadin score in the last inning is 39.

10 (7) b ID : ae-9-data-handling-probability-statistics [10] We know that, the integers divisible by and 3 are also divisible by 6. Theref ore, the number of integers divisible by and not divisible by 3 = The number of integers divisible by - The number of integers divisible by 6 The number of integers in the set 1,, 3,...116, divisible by = quotient of 116/ = 58 The number of integers in the set 1,, 3,...116, divisible by 6 = quotient of 116/6 = 19 Now, the number of integers in the set 1,, 3,...116, divisible by and not divisible by 3 = = 39 The probability that an integer in the set 1,, 3, is divisible by and not divisible by 3 The integers in the set 1,, 3, is divisible by and not divisible by 3 = The total number of integers in the set 1,, 3, = (8) d The number of possible outcomes when 6 coins are tossed is 6 To f ind out the probability of getting at least one tails, let's look at the outcomes where this is not true i.e. the number of outcomes where you do not have even one tails Obviously, this is the case where you have all the tosses giving heads There is only one case where you can get all heads So the probability of getting at least one tails = = 63 64

11 (9) b. 37 ID : ae-9-data-handling-probability-statistics [11] If you look at the question caref ully, you will notice that the age of the employees was as f ollows 41, 36, 47, 3, 48, 41, 7, 31, 6, 35, 43 Sum of age of the employees = = 407 Total number of employees = 11 Average age of the employees = Sum of age of the employees T otal number of employees = = 37 Now the average age of the employees is 37. (10) c. 69 If the average marks of 0 students was 71, then Sum of all the marks of the students = 0 x 71 = 140 To f ind out the average of the other 18 students, we subtract the marks of Givon and Kamal Total marks of other 18 students = = 14 The average marks of the remaining students = = 69

12 (11) a ID : ae-9-data-handling-probability-statistics [1] First we need to f ind the total number of tyres that are given here. We add the number of tyres = 000. To f ind the probability that the tyre Anan purchased would last more than 0000 kilometers, we need to add the number of tyres that lasted more than 0000 kilometers. This is = Step 4 The probability that the tyre lasts more than 0000 km = (1) d. 7 There are 366 days in a leap year If we divide 366 by 7 (since there are seven days in a week), we will get an answer of 5, with a remainder of This means that a leap year will have 5 Sundays, 5 Mondays, 5 Tuesdays, 5 Wednesdays, 5 T hursdays, 5 Fridays and 5 Saturdays. Apart f rom these there will be two other days. This means that there will be two weekdays that occur 53 times. T he two days could be (Sunday, Monday), (Monday, T uesday), (T uesday, Wednesday), (Wednesday, T hursday), (T hursday, Friday), (Friday, Saturday), or (Saturday, Sunday) - a total of seven combinations Step 4 Out of these seven combinations, two of them have a Sunday Step 5 So the probability of either of those two days being a Sunday is 7

13 (13) even ID : ae-9-data-handling-probability-statistics [13] 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 = = 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. (14) 1 9 To f ind the average, let's take all the ages in terms of months. We get, Aziz's age = 1 x = 148 months old. Kalil's age = 1 x = 147 months old. Ghassan's age = 13 x x 1 = 165 months old. Hadya's age = 1 x x 1 = 150 months old. To get the average, we add these ages and divide by The average age in months = 4 Converting 153 months to years, we get, The Average age = 153/1 = 1 years and 9 months. = 61 4 = 153 months

14 (15) 5.5 ID : ae-9-data-handling-probability-statistics [14] Marks T ally Marks Number of students If you look at the given table f rom top to bottom caref ully, you will notice that the marks obtained by 30 students in German are arranged in ascending order. Total number of students are 30 which is even. So, median is equal to the average of ( n ) th and ( n +1) th student's marks in German, (where n is the number of students) ( n ) th = ( 30 ) th = 15 th ( n +1) th = ( ) th = 16 th If you count the number of students in tally marks column of the table, you will notice that the marks obtained by 15 th and 16 th students in German are 5 and 6 respectively median = = 11 = 5.5 Now the median of their scores is 5.5.