Grade 10 Mean, Mode and Median

Transcription

1 ID : ae-10-mean-mode-and-median [1] Grade 10 Mean, Mode and Median For more such worksheets visit Answer the questions (1) A poll is taken among people working in a town. The aim was to see what their annual salaries were. Annual Salary Number of people Less than Dhs Dhs40001 to Dhs Dhs75001 to Dhs Dhs to Dhs More than Dhs If you choose a person at random from this group, what is the probability that he or she earns more than Dhs40000 annually? (2) There are 4 pens belonging to 4 students. The pens were put into a box, and each student pulls out an pen one after the other. What is the probability that each student gets his or her own pen? (3) If the mean of the following data is 38.3, and the sum of x and y is 24, then what is the value of y. Class interval Frequency x y (4) Find the median of the following data Class interval Frequency (5) If the mean of the following data is 22.6, find the value of x. Class interval Frequency x 20

2 (6) The table below shows the number of books read by five children in a year. Name Number of books Hadya 25 Aden 12 Haroun x Aisha 27 Aiesha 29 ID : ae-10-mean-mode-and-median [2] If the average number of books read by the children was 25, then how many books did Haroun read over the year? (7) If a prime number is less than 31, what is the probability that it is also less than 23. (8) If the median value of the data below is 60, what is the value of x Class interval Frequency x 17 (9) A tyre manufacturer keeps the record of how much distance the tyres manufactured by the company travel before failing. They find the following data Distance traveled in kilometers Number of failing tyres Less than to to to More than If Aziza buys a tyre from them, what is the probability that it will last more than kilometers? (10) Two dice are rolled. What is the probability that the two numbers add up to a prime number? (11) 5 coins are tossed in parallel. What is the probability of getting at least one tail? (12) Adnan draws one card from a shuffled deck of 52 cards. What is the probability that he drew a spade or a King? (13) The letters of the word PROBABILITY are rearranged in a random order. What is the probability that the letters P and L have exactly 3 letters between them? (14) From a deck of cards, Ameerah took out one card at random. What is the probability that she got a prime number? (15) Find the mean of the following data (Hint: Use the Step-Deviation method). Class interval Frequency

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4 Answers ID : ae-10-mean-mode-and-median [4] (1) First we need to find the total number of people among whom the poll was taken. We add the number of people in the various sets = To find the probability that the the random person chosen has a salary of more than Dhs40000, we need to add the number of people who have salaries greater than this number. This is = Step 4 So, 9840 people out of a total of earn an annual salary greater than Dhs The probability that the randomly chosen person has a salary greater than Dhs40000 = (2) 1 24 We have 4 pens and 4 students. Let's first see in how many different ways can the pens be distributed among the students We know that 4 objects can be distributed in 4! = 4 x 3 x...x 1 = 24 ways Out of these 24 ways, there is only one distribution where each student got his or her own pen Step 4 So the probability of each student getting his or her own pen = 1 24

5 (3) 8 ID : ae-10-mean-mode-and-median [5] Data can be re-arranged as shown in following table, Class interval Frequency (f i ) Class mean ((x i )) f i x i x 35 35x y 45 45y Total Σf i = 76 + x + y Σf i x i = x + 45y Now, mean can be calculated as following, m x = (Σf i x i )/(Σf i ) m x = x + 45y 76 + x + y It is given that, x + y = 24, or x = 24 - y, therefore m x = (24 - y) + 45y y m x = m x = y Step 4 It is given that mean m x = 38.3, therefore = y y = -10 y = 8

6 (4) 131 ID : ae-10-mean-mode-and-median [6] The data can be re-arranged as shown in following table, Class interval Frequency(f i ) Cumulative frequency(cf) = = = = 68 From the given table we notice that, n = 68 and n/2 = 34. The Cumulative frequency(cf) just greater than or equal to the n/2 is 33, belonging to the interval Therefore, the median class = , Lower limit(l) of the median class = 120, Class size(h) = 10, Frequency(f) of the median class = 10, Cumulative frequency(cf) of the class preceding median class = 23 The median = l + ( n/2 - cf f ) h = ( = ) 10 Thus, the median of the data is 131.

7 (5) 12 ID : ae-10-mean-mode-and-median [7] Data can be re-arranged as shown in following table, Class interval Frequency (f i ) Class mean ((x i )) f i x i x 35 35x Total Σf i = 88 + x Σf i x i = x Now, mean can be calculated as following, m x = Σf i x i /Σf i m x = x 88 + x It is given that, the mean of the data is 22.6, therefore, x 22.6 = 88 + x 22.6(88 + x) = x x = x 22.6x - 35x = x = x = x = 12

8 (6) 32 ID : ae-10-mean-mode-and-median [8] Lets assume, Haroun read x number of books Aaverage number of books read by five children, N av = ( x )/5 = (x + 93)/5 Since it is given that N av = 25, (x + 93)/5 = 25 x + 93 = 25 5 x + 93 = 125 x = x = 32 (7) 4 5 There are 10 prime numbers i.e. (2, 3, 5, 7, 11, 13, 17, 19, 23, 29) which are less than 31. Out of these 8 prime numbers are less than 23. Probability (The prime number is less than 23) = Number of prime numbers less than 23 Number of prime numbers less than 31 = 8 10 = 4 5

9 (8) 11 ID : ae-10-mean-mode-and-median [9] The data can be re-arranged as shown in following table, Class interval Frequency(f i ) Cumulative frequency(cf) = = x 44 + x = 44 + x x + 17 = 61 + x From the given table we notice that, n = 61 + x and n/2 = (61 + x)/2 = 61/2 + x/2. The Cumulative frequency(cf) just greater than or equal to the n/2 is 44, belonging to the interval Therefore, the median class = 56-64, Lower limit(l) of the median class = 56, Class size(h) = 8, Frequency(f) of the median class = 16, Cumulative frequency(cf) of the class preceding median class = 28. The median = l + ( n/2 - cf f ) h 60 = 56 + ( (61 + x)/2 - (28) 16 ) 8 On solving above equatoin, we get, x = 11

10 ID : ae-10-mean-mode-and-median [10] (9) First we need to find the total number of tyres that are given here. We add the number of tyres = To find the probability that the tyre Aziza purchased would last more than kilometers, we need to add the number of tyres that lasted more than kilometers. This is = Step 4 The probability that the tyre lasts more than km = (10) 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 from the list of possible values from each Dice, we get numbers ranging from 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 = 15 possible ways to get a prime number So the probability of getting the two numbers add up to a prime is 15 36

11 ID : ae-10-mean-mode-and-median [11] (11) The number of possible outcomes when 5 coins are tossed is 2 5. To find out the probability of getting at least one tail, let's look at the outcomes where this is not true. i.e. the number of outcomes where you do not have even one tail. Obviously, this is the case where you have all the tosses giving head. There is only one case where you can get all head. 1 So, the probability of getting at least one tail = 1-32 =

12 ID : ae-10-mean-mode-and-median [12] (12) 4 13 We know that there are 52 cards in a deck out of which 13 cards are of spade. The probability that he drew a spade is = 1 4. There are 4 cards of King in a deck. So, the probability that he drew a King is 4 52 = We need to add these two probabilities as we are asked for the probability of getting a spade or a King. This gives us = Step 4 But,there is another thing to keep track of. In our list, we have counted the King of spade twice, - once as a spade in the first component of the sum above, - and second as a King in the second component above. We should only count it once, so we remove one of the times we count it. The probability of a specific card being chosen is 1, so we subtract that to get the result. 52 Step 5 The probability that he drew a spade or a King = = 4 13.

13 ID : ae-10-mean-mode-and-median [13] (13) The first step is finding out all the possible arrangements of the letters of the word PROBABILITY. Note that we are not interested in "unique" arrangements (in some kind of problems we would be), but just the total possible arrangements. There are 11 letters in the word PROBABILITY. So, they can be arranged in 11! ways i.e ways. To see this, if you pick any letter, it can be placed in any of the possible 11 positions. Now, the second letter can be placed in any of the other 10 positions (11 minus the one taken up by the first letter). So, the two letters can be arranged in ways. Similarly we can arrange the remaining letters. Now, we know that the total number of possible arrangements = 11! We need to look for the possible arrangements where the letters P and L have exactly 3 letters between them. Step 4 This can be seen by inspection. If P is at the first position (first letter of the word), then L will be at position 5 (since there are 3 letters between them). Similarly, if P is at the second position, then L will be at position 6. There are 7 such possible arrangements, the last one having P at the 7 th position and L at the last position. Step 5 Interchanging the positions of L and P in all the cases given in the above step, we get 7 possible arrangements. So, we can place the letters P and L in 14 ways. Now, we have filled in 2 of the 11 positions with P and L in 14 ways. Since there is 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 P and L have exactly 3 letters between them is 14 9!. Step 6 Now, we can work out the probability of rearranging the letters of the word PROBABILITY such that P and L have exactly 3 letters between them. P (arrangement) = Number of possible arrangements where P and L have 3 letters between them 14 9! = = Total number of possible arrangements of the letters of the word PROBABILITY 11!

14 ID : ae-10-mean-mode-and-median [14] (14) 4 13 There are 52 cards in a deck, with 13 cards of each suit (diamond, club, heart, and spade). 13 cards of a suit are the numbers from 2 to 10, 3 face cards and 1 ace card. From the numbers 2 to 10, only 2, 3, 5, and 7 are prime. So there are four prime numbered cards per suit, which gives a total of 16 prime numbered cards. Step 4 So, there are 16 prime numbered cards out of a total of 52 cards. Number of prime numbered cards So, the probability that we pick a prime numbered card = Total number of cards = = 4 13

15 (15) ID : ae-10-mean-mode-and-median [15] Class interval Frequency 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 following table, Class interval Frequency(f i ) Class mean(x i ) d i = x i u i = d i /10 f i u i Total Σf i = 100 Σf i u i = 6 Now, mean can be calculated by the Step-Deviation method, as following, m x = a + ( = ( = Σf i u i Σf i ) h ) 10 The mean of the data is