ID : pk-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) 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 11 6 23 20 36 (2) In 1996, 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? (3) 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 7 10 14 10 23 (4) There are a total of 16 chocolates - 4 each in the f lavors of orange, banana, cof f ee and strawberry. There are also 5 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? (5) The numbers 1 to 11 are written on 11 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 (6) If the mean of the f ollowing data is 22.8, f ind the value of x. Class interval 0-10 10-20 20-30 30-40 40-50 Frequency x 24 12 14 22 a. 31 b. 29 c. 26 d. 28
ID : pk-10-mean-mode-and-median [2] (7) 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 320 10001 to 20000 320 20001 to 40000 528 More than 40000 792 If Altaf buys a tyre f rom them, what is the probability that it will last more than 10000 kilometers? a. 1614 2000 c. 1640 2000 b. 1630 2000 d. 1653 2000 (8) The letters of the word STATISTICAL are rearranged in a random order. What is the probability that the letters have exactly 4 letters between them? a. 1 b. 6 105 c. 12 110 d. 8 110 (9) Find the mean of the f ollowing data (Hint: Use the Step-Deviation method). Class interval 200-210 210-220 220-230 230-240 240-250 Frequency 18 20 20 16 26 a. 231.2 b. 236.2 c. 216.2 d. 226.2 (10) Maleeha is participating in a race. The probability that she will come f irst in the race is 0.3. The probability that she will come second in the race is 0.25. The probability that she will come in 3rd is 0.35, and the probability that she will be 4th is 0.1. What is the probability that she will win 2nd position or better in the race? a. 0.45 b. 0.6 c. 0.65 d. 0.55 (11) A bag contains 2 red balls, 5 blue balls, and 8 green balls. Tasneem draws 2 balls out of the bag. What is the probability that she gets a green ball and a blue ball? a. 42 112 b. 38 105 c. 40 105 d. 42 105
(12) If the median value of the data below is 75, what is the value of x Class interval 60-66 66-72 72-78 78-84 84-90 Frequency 20 10 12 x 12 ID : pk-10-mean-mode-and-median [3] a. 17 b. 18 c. 20 d. 19 (13) A card is drawn f rom a shuf f led deck of 52 cards. What is the probability of getting a card greater than 3 but lesser than 10? a. 24 52 b. 12 52 c. 6 52 d. 18 52 (14) The table below shows the number of books read by f ive children in a year. Name Number of books Saif 26 Shabana 22 Mehek x Azhar 28 Masood 33 If the average number of books read by the children was 25, then how many books did Mehek read over the year? a. 16 b. 17 c. 13 d. 12 (15) A box contains 6 black and 6 yellow balls. Fahad takes out a ball, puts it back, and then takes out another ball. What is the probability that both balls were black? a. 6 144 b. 6 12 c. 36 144 d. 36 72 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : pk-10-mean-mode-and-median [4] (1) 115.2 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 11 11 96-104 6 11 + 6 = 17 104-112 23 17 + 23 = 40 112-120 20 40 + 20 = 60 120-128 36 60 + 36 = 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 60, 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 = 40 The median = l + ( n/2 - cf f ) h = 112 + ( = 115.2 48-40 20 ) 8 Thus, the median IQ of the students is 115.2.
(2) 307 ID : pk-10-mean-mode-and-median [5] 366 The key thing to note here is that 1996 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
(3) 165.5 ID : pk-10-mean-mode-and-median [6] The data can be re-arranged as shown in f ollowing table, Class interval Frequency(f i ) Cumulative frequency(cf) 150-155 cm 7 7 155-160 cm 10 7 + 10 = 17 160-165 cm 14 17 + 14 = 31 165-170 cm 10 31 + 10 = 41 170-175 cm 23 41 + 23 = 64 From the given table we notice that, n = 64 and n/2 = 32. The Cumulative f requency(cf ) just greater than or equal to the n/2 is 41, 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 = 31. The median = l + ( n/2 - cf f ) h = 165 + ( = 165.5 32-31 10 ) 5 Thus, the median height of the data is 165.5.
(4) 255 ID : pk-10-mean-mode-and-median [7] 256 There are 5 children. They each have some f avourite f lavor f rom among orange, banana, cof f ee and strawberry. Each child's f avourite could be any one of the 4 choices The possible combinations of f lavors they like = 4 x 4 x 4 x 4 x 4 = 1024 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 4 f lavors in each choice, and 5 children. The only way f or a child not to get his or her f avourite is if all 5 children choose the same f lavor. This is because if even one child chooses some other f lavor f rom the rest, the other 4 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 4 f lavors, this can happen in 4 cases 4 So the probability that a child does not get his or her f lavor = 1024 Step 4 Theref ore the probability that all children get their choice = 1-4 1024 = 255 256
(5) 15 ID : pk-10-mean-mode-and-median [8] 99 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 = 11, so n=5 The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between 1 and 11 is the ratio of - Number of ways we can get an A.P f rom 3 random numbers between 1 to 11, and - Number of ways to select 3 random numbers between 1 to 11 Step 4 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 11 = n(4n 2-1) 3 Here n = 5, so we can simplif y it as 165 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 (11-2), 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 : pk-10-mean-mode-and-median [9] 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=5, so this is 25 0 So the probability is 15 99
(6) d. 28 ID : pk-10-mean-mode-and-median [10] Data can be re-arranged as shown in f ollowing table, Class interval Frequency (f i ) Class mean ((x i )) f i x i 0-10 x 5 5x 10-20 24 15 360 20-30 12 25 300 30-40 14 35 490 40-50 22 45 990 Total Σf i = 72 + x Σf i x i = 2140 + 5x Now, mean can be calculated as f ollowing, m x = Σf i x i /Σf i m x = 2140 + 5x 72 + x It is given that, the mean of the data is 22.8, theref ore, 2140 + 5x 22.8 = 72 + x 22.8(72 + x) = 2140 + 5x 1641.6 + 22.8x = 2140 + 5x 22.8x - 5x = 2140-1641.6 17.8x = 498.4 x = 498.4 17.8 x = 28
(7) c. 1640 2000 First we need to f ind the total number of tyres that are given here. We add the number of tyres 40 + 320 + 320 + 528 + 792 = 2000. ID : pk-10-mean-mode-and-median [11] To f ind the probability that the tyre Altaf purchased would last more than 10000 kilometers, we need to add the number of tyres that lasted more than 10000 kilometers. This is 320 + 528 + 792 = 1640. Step 4 The probability that the tyre lasts more than 10000 km = 1640 2000.
(8) c. 12 110 ID : pk-10-mean-mode-and-median [12] The f irst step is f inding out all the possible arrangements of the letters of the word STATISTICAL. Note that we are not interested in "unique" arrangements (in some kind of problems we would be), but just the total possible arrangement The answer to this is that there are 11 letters in this word, and theref ore the letters can be arranged in 11! ways i.e. 11x10x9x...2x1 ways To see this, take any of the letters - it can be in any of the possible 11 positions. For each of these position, a second letter can be in any of the other 10 positions (11 minus the one taken up by the f irst letter) So the two letters can appear in 11x10 combinations For each of these 110 combinations, a third letter can be in any of the remaining 9 places, and so on Now we know the total number of arrangements (11!) possible, and need to look f or the possible arrangements where the letters C and L have exactly 4 letters between them) Step 4 This can be seen by inspection If C is in the f irst position (f irst letter of the word), then L would need to be in position 6 (since there are 4 letters between them Similarly, if C is in the second position, then L would need to be in position 7 There are 6 such positions, the last one having C in the 6th position and L being in the last position Step 5 The same thing can be seen with L being bef ore C. There are 6 such positions So the total number of positions f or the two letters where this condition is met is 12 Now we have f illed in 2 of the 11 letters with C and L in 12 ways The remaining 9 letters can take any of the remaining 9 positions f or each of these Since there are no restriction on the remaining 9 letters, the number of possible arrangements of 9 letters in 9 positions is 9! So the total ways to rearrange all the letters so that C and L have exactly 4 letters between them is 12 x 9! Step 6 Now we can work out the probability of rearranging the letters of STATISTICAL so that C and L have exactly 4 letters between them Arrangements where the two letters have 4 letters between them P (arrangement) = = Total possible arrangements of the letters of STATISTICAL 12 x 9! 11! = 12 110
ID : pk-10-mean-mode-and-median [13] (9) d. 226.2 Class interval 200-210 210-220 220-230 230-240 240-250 Frequency 18 20 20 16 26 If we look at the table, we notice that, class size(h) = 10. Taking 225 as a mean(a), the data can be re-arranged as shown in f ollowing table, Class interval Frequency(f i ) Class mean(x i ) d i = x i - 225 u i = d i /10 f i u i 200-210 18 205-20 -2-36 210-220 20 215-10 -1-20 220-230 20 225 0 0 0 230-240 16 235 10 1 16 240-250 26 245 20 2 52 Total Σf i = 100 Σf i u i = 12 Now, mean can be calculated by the Step-Deviation method, as f ollowing, m x = a + ( Σf i u i Σf i ) h = 225 + ( 12 = 226.2 100 ) 10 The mean of the data is 226.2.
(11) c. 40 105 ID : pk-10-mean-mode-and-median [14] There are a total of 2+5+8 = 15 balls in the bag The number of ways to pick out 2 balls f rom a set of 15 balls is 15 C2 = 15 x (15-1)/2 = 105 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 8 x 5 = 40 Step 4 The probability that she gets a green ball and a blue ball is theref ore 40 105
(12) b. 18 ID : pk-10-mean-mode-and-median [15] The data can be re-arranged as shown in f ollowing table, Class interval Frequency(f i ) Cumulative frequency(cf) 60-66 20 20 66-72 10 20 + 10 = 30 72-78 12 30 + 12 = 42 78-84 x 42 + x = 42 + x 84-90 12 42 + x + 12 = 54 + x From the given table we notice that, n = 54 + x and n/2 = (54 + x)/2 = 54/2 + x/2. The Cumulative f requency(cf ) just greater than or equal to the n/2 is 42, belonging to the interval 72-78. Theref ore, the median class = 72-78, Lower limit(l) of the median class = 72, Class size(h) = 6, Frequency(f ) of the median class = 12, Cumulative f requency(cf ) of the class preceding median class = 30. The median = l + ( n/2 - cf f ) h 75 = 72 + ( (54 + x)/2 - (30) 12 ) 6 On solving above equatoin, we get, x = 18
(13) a. 24 52 ID : pk-10-mean-mode-and-median [16] To f igure this out, we identif y how many cards that match this condition are present in the deck There are f our suits, and each suit has 13 cards : 1 to 9, J, Q, K and A The cards that are > 3 but < 10 in each suit are 4,...,9 The number of such cards is 6 Step 4 Since there are 4 such suits, the total probability is 4 x 6 52 = 24 52 (14) a. 16 Lets assume, Mehek read x number of books Aaverage number of books read by f ive children, N av = (26 + 22 + x + 16 + 28 + 33)/5 = (x + 109)/5 Since it is given that N av = 25, (x + 109)/5 = 25 x + 109 = 25 5 x + 109 = 125 x = 125-109 x = 16
(15) c. 36 144 ID : pk-10-mean-mode-and-median [17] The total number of balls in the box is 12 The event we are looking f or is that both the balls picked by Fahad are black The key thing here is that Fahad 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 Step 4 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 black ball in one pick is 6 12 Step 6 The probability of getting black balls in both picks is 6 12 x 6 12 = 36 144