Grade 10 Mean, Mode and Median

ID : ww-10-mean-mode-and-median [1] Grade 10 Mean, Mode and Median For more such worksheets visit www.edugain.com Answer t he quest ions (1) What is the probability that a leap year will contain 53 Wednesdays? (2) What is the probability that a leap year will contain 53 Sundays and 52 Mondays? (3) A bag contains 6 red balls, 4 blue balls, and 6 green balls. Kimberly draws 2 balls out of the bag. What is the probability that she gets a green ball and a blue ball? (4) John selects 3 numbers randomly f rom the f ollowing set of 5 numbers 8, 2, 5, 6 and 9. 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 14 15? (5) Find the median height f rom the f ollowing data Class interval 150-155 cm 155-160 cm 160-165 cm 165-170 cm 170-175 cm Frequency 9 6 12 10 21 (6) A box contains 4 green and 3 red balls. Michael takes out a ball, puts it back, and then takes out another ball. What is the probability that both balls were green? (7) Two dice are rolled. What is the probability that the two numbers add up to a prime number? (8) If a prime number is less than 29, what is the probability that it is also less than 17. (9) The numbers 1 to 15 are written on 15 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? Choose correct answer(s) f rom given choice (10) The distribution of IQ among a set of students is given below. What is their median IQ? Class interval 88-96 96-104 104-112 112-120 120-128 Frequency 7 8 21 20 40 a. 126.8 b. 121.8 c. 116.8 d. 111.8

(11) 5 coins are tossed in parallel. What is the probability of getting at least one heads? ID : ww-10-mean-mode-and-median [2] a. 31 32 b. 29 32 c. 1 32 (12) There are a total of 9 chocolates - 3 each in the f lavors of grape, cherry and banana. There are also 4 children. If each child is allowed to choose their own f avorite f lavor, what is the probability that all of them will get f lavors of their choice? a. 3 27 b. 26 27 c. 25 27 d. 25 28 (13) 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 5000 40 5000 to 10000 240 10001 to 20000 200 20001 to 40000 608 More than 40000 912 If Karen buys a tyre f rom them, what is the probability that it will last more than 20000 kilometers? a. 1520 2000 c. 1483 2000 b. 1486 2000 d. 1501 2000

ID : ww-10-mean-mode-and-median [3] (14) The table below shows the number of books read by f ive children in a year. Name Number of books Jason 31 Sarah 39 Paul x Steven 23 Lisa 33 If the average number of books read by the children was 32, then how many books did Paul read over the year? a. 34 b. 39 c. 36 d. 31 (15) Steven draws 4 cards out of a deck of 52 cards. What is the probability that he draws a Queen, a 9, a King and a 7? a. 64 270725 b. 256 270725 c. 3 52 d. 4 52 2016 Edugain (www.edugain.com). Many more such worksheets can be All Rights Reserved generated at www.edugain.com

Answers ID : ww-10-mean-mode-and-median [4] (1) 2 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 52, with a remainder of 2 This means that a leap year will have 52 Sundays, 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 T hursdays, 52 Fridays and 52 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 Out of these seven combinations, two of them have a Wednesday Step 5 So the probability of either of those two days being a Wednesday is 2 7

ID : ww-10-mean-mode-and-median [5] (2) 1 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 52, with a remainder of 2 This means that a leap year will have 52 Sundays, 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 T hursdays, 52 Fridays and 52 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 Out of these seven combinations, only one of them has Sunday but no Monday Step 5 So the probability of either of those two days being a Sunday is 1 7 (3) 24 120 There are a total of 6+4+6 = 16 balls in the bag The number of ways to pick out 2 balls f rom a set of 16 balls is 16 C2 = 16 x (16-1)/2 = 120 The number of ways to pick a green ball and a blue ball is obtained by multiplying the number of the balls of each of these colors. This is 6 x 4 = 24 The probability that she gets a green ball and a blue ball is theref ore 24 120

(4) 24 ID : ww-10-mean-mode-and-median [6] 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 2, 5, 6, 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 Now we need to f igure out how many of these 30 f ractions are greater than 2 14 15 In gt, we see that the whole number is 2, the numerator is 14 and the denominator is 15 We see that the numerator 14 is larger than the largest number in the set of numbers given to us, and the denominator 15 is one larger than 14 The implication of this is that 2 14 will be larger than any f raction that can be f ormed 15 f rom the set of numbers 2, 5, 6, 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 5, 6, 8 and 9 Step 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 ID : ww-10-mean-mode-and-median [7] Step 6 So using each of the numbers f rom 5, 6, 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 5, 6, 8 and 9 as the whole number = 6 x 4 = 24 Step 7 Out of the 30 f ractions, 24 will be greater than 2 14 15 Step 8 The probability is theref ore = 24 30

(5) 166 ID : ww-10-mean-mode-and-median [8] The data can be re-arranged as shown in f ollowing table, Class interval Frequency(f i ) Cumulative frequency(cf) 150-155 cm 9 9 155-160 cm 6 9 + 6 = 15 160-165 cm 12 15 + 12 = 27 165-170 cm 10 27 + 10 = 37 170-175 cm 21 37 + 21 = 58 From the given table we notice that, n = 58 and n/2 = 29. The Cumulative f requency(cf ) just greater than or equal to the n/2 is 37, belonging to the interval 165-170 cm. Theref ore, the median class = 165-170 cm, Lower limit(l) of the median class = 165, Class size(h) = 5, Frequency(f ) of the median class = 10, Cumulative f requency(cf ) of the class preceding median class = 27. The median = l + ( n/2 - cf f ) h = 165 + ( = 166 29-27 10 ) 5 Thus, the median height of the data is 166.

(6) 16 ID : ww-10-mean-mode-and-median [9] 49 The total number of balls in the box is 7 The event we are looking f or is that both the balls picked by Michael are green The key thing here is that Michael picks a ball, looks at the color, and puts it back in the box This means the box is the same in both cases f or both picks - it has the same number of balls in both cases This also means that the two events - picking up of ball 1 and 2 - are independent events, and theref ore their probabilities can be multiplied to get the f inal probability of both events happening. Step 5 The probability of getting a green ball in one pick is 4 7 Step 6 The probability of getting green balls in both picks is 4 7 x 4 7 = 16 49

(7) 15 ID : ww-10-mean-mode-and-median [10] 36 The two dice that are rolled can show any of these values Dice 1 : 1, 2, 3, 4, 5, 6 Dice 2 : 1, 2, 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 2 (when both Dice show 1) to 12 (when both dice show 6). Let's enumerate the prime numbers between 2 and 12. They are 2, 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 2-2: The only way to get this is when we roll (1,1). 1 possibility - 3: We can get this by (1,2) or (2,1). 2 possibilities. - 5: We can get this by (2,3), (3,2), (1,4) or (4,1). 4 possibilities. - 7: We can get this by (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 possibilities. - 11: We can get this by (5,6), or (6,5). 2 Possibilities This gives us a total of 1 + 2 + 4 + 6 + 2 = 15 possible ways to get a prime number So the probability of getting the two numbers add up to a prime is 15 36 (8) 2/3 There are 9 prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23) which are less than 29 Of these 6 are less than 17 Theref ore probability is 6/9 = 2/3

(9) 21 ID : ww-10-mean-mode-and-median [11] 195 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 (2n+1) Here, we can see f rom the equation 2n+1 = 15, so n=7 The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between 1 and 15 is the ratio of - Number of ways we can get an A.P f rom 3 random numbers between 1 to 15, and - Number of ways to select 3 random numbers between 1 to 15 Let's look at the second part f irst. Three tickets can be drawn f rom (2n+1) numbers is in [(2 x n) + 1] C 3 ways i.e. Number of ways 3 tickets can be drawn = (2n+1)(2n)(2n-1) 3x2x1 Simplif ying this, we get the number of ways to draw 3 numbers between 1 and 15 = n(4n 2-1) 3 Here n = 7, so we can simplif y it as 455 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. 1,2,3 or 3,5,7 etc. They are in the f orm (a, a+d, a + 2d), where a is an integer f rom 1 to (15-2), 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 2, and so on Step 6 So what are the possible sequences with interval 1. They are (1,2,3) (2,3,4)... (2n-1,2n,2n+1) T here are theref ore 2n-1 such possible sequences Step 7 Similarly, let's look at A.P with interval 2 between the terms. They are (1,3,5) (2,4,6)

...<2n-3,2n-1,2n+1> T here are 2n-3 such possible sequences ID : ww-10-mean-mode-and-median [12] Step 8 We can generalize this to say that the number of such sequences with interval 'd' is (2n- (2d-1)) Obviously the largest possible integer is d=n, with just one sequence (1, n+1, 2n+1) Step 9 So the total number of such sequences is (2n-1) + (2n-3) + (2n-5) +... + 5 + 3 + 1 This is itself an AP with n terms and d=2 The sum of this sequence is n 2 [2 + (n-1)2] Simplif ying, we get n 2 Here n=7, so this is 49 0 So the probability is 21 195

(10) c. 116.8 ID : ww-10-mean-mode-and-median [13] The distribution of IQ among the set of students can be re-arranged as shown in f ollowing table, Class interval Frequency(f i ) Cumulative frequency(cf) 88-96 7 7 96-104 8 7 + 8 = 15 104-112 21 15 + 21 = 36 112-120 20 36 + 20 = 56 120-128 40 56 + 40 = 96 From the given table we notice that, n = 96 and n/2 = 48. The Cumulative f requency(cf ) just greater than or equal to the n/2 is 56, belonging to the interval 112-120. Theref ore, the median class = 112-120, Lower limit(l) of the median class = 112, Class size(h) = 8, Frequency(f ) of the median class = 20, Cumulative f requency(cf ) of the class preceding median class = 36 The median = l + ( n/2 - cf f ) h = 112 + ( = 116.8 48-36 20 ) 8 Thus, the median IQ of the students is 116.8.

(11) a. 31 32 ID : ww-10-mean-mode-and-median [14] The number of possible outcomes when 5 coins are tossed is 2 5 To f ind out the probability of getting at least one heads, let's look at the outcomes where this is not true i.e. the number of outcomes where you do not have even one heads Obviously, this is the case where you have all the tosses giving tails There is only one case where you can get all tails So the probability of getting at least one heads = 1-1 32 = 31 32 (12) b. 26 27 There are 4 children. They each have some f avourite f lavor f rom among grape, cherry and banana. Each child's f avourite could be any one of the 3 choices The possible combinations of f lavors they like = 3 x 3 x 3 x 3 = 81 Let's now consider the options where even one child, no matter what his or her f avourite f lavour is, does not get his or her choice Now there are 3 f lavors in each choice, and 4 children. The only way f or a child not to get his or her f avourite is if all 4 children choose the same f lavor. This is because if even one child chooses some other f lavor f rom the rest, the other 3 children could get their f avourite, no matter what they choose The cases where some child might not get his or her choice is theref ore when they all choose the same f lavor Since there are 3 f lavors, this can happen in 3 cases 3 So the probability that a child does not get his or her f lavor = 81 Theref ore the probability that all children get their choice = 1-3 81 = 26 27

(13) a. 1520 2000 ID : ww-10-mean-mode-and-median [15] First we need to f ind the total number of tyres that are given here. We add the number of tyres 40 + 240 + 200 + 608 + 912 = 2000. To f ind the probability that the tyre Karen purchased would last more than 20000 kilometers, we need to add the number of tyres that lasted more than 20000 kilometers. This is 608 + 912 = 1520. The probability that the tyre lasts more than 20000 km = 1520 2000. (14) a. 34 Lets assume, Paul read x number of books Aaverage number of books read by f ive children, N av = (31 + 39 + x + 34 + 23 + 33)/5 = (x + 126)/5 Since it is given that N av = 32, (x + 126)/5 = 32 x + 126 = 32 5 x + 126 = 160 x = 160-126 x = 34

(15) b. 256 270725 ID : ww-10-mean-mode-and-median [16] There are 4 Queens in a deck of cards. At this point there are also 52 cards in the deck There are 4 9s in a deck of cards. Now there are 51 cards remaining in the deck There are 4 Kings in a deck of cards. Now there are 50 cards remaining in the deck There are 4 7s in a deck of cards. Now there are 49 cards remaining in the deck Step 5 You can select 4 cards without replacement f rom a deck in 52 C 4 ways = 52 x 51 x 50 x 49 4 x 3 x 2 x 1 Step 6 4 x 4 x 4 x 4 So the answer is 52 x 51 x 50 x 49 4 x 3 x 2 x 1 Step 7 Simplif ying we get 256 270725