Statistics 2013

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<ul><li><p>additionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticszefryadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticsadditionalmathematicsstatisticshjklzxcvbnmqwert</p><p>STATISTICS </p><p> Name </p><p> ........................................................................................ </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>CHAPTER 7 : STATISTICS </p><p>7.1 MEASURES OF CENTRAL TENDENCY (i) Ungrouped data data that is not grouped into classes (ii) Grouped data - data that is grouped in certain classes </p><p>7.1.1 Calculating the mean of ungrouped data </p><p> N</p><p>xx</p><p>Example 1: </p><p>Find the mean of 58, 67, 45, 73 and 77 </p><p>Example 2: </p><p>Number 1 2 3 4 5 </p><p>Frequency 5 8 4 6 2 </p><p> Find the mean of the number. </p><p>7.1.2 Determining mode of ungrouped data </p><p>Mode is the value which appears the most number of times in a set of data (value that has the </p><p>highest frequency) </p><p>Example 1: Determine the mode for the following sets of number. </p><p> (a) 2, 5, 6, 2, 6, 7, 2, 4, 8, 2 Answer: .. </p><p> (b) 5, 5, 8, 10, 4, 4 Answer: .. </p><p> (c) 2, 2, 3, 4, 4 , 4, 6, 6 Answer: .. </p><p>Note: It is possible that a set of data either has more than one mode or has no mode. </p><p> Where </p><p>x - mean of the set data. x values in the set of data N total number of data </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Example 2: Determine the mode of the number of pens a student has. </p><p>Number of pens 1 2 3 4 5 </p><p>Number of students 6 7 5 3 3 </p><p>Answer: .. </p><p>7.1.3 Determining the median of ungrouped data </p><p>When the values in set of data are arranged in either ascending or descending order, the value that </p><p>lies in the middle is the median. </p><p>Example 1: 3, 4, 5, 6, 7, 8, 9 </p><p>Example 2: 21, 20, 19, 18, 17, 16, 15, 14 </p><p>Activity 1: </p><p>1. Calculate the mean, mode and median for the following sets of data. </p><p> (i) 1, 4, 5, 8, 9, 8, 8, 7, 4 (ii) 5, 8, 12, 10, 5, 3, 7, 5, 20, 10 </p><p>2. (i) Find the mean of 6, 8, 4, 9 and 11 </p><p> (ii) Find the value of x if the mean of 4, 5, 6, 7, 11 and x is 7. </p><p>3. Find the mode of each following sets of data. </p><p>(i) 8, 6, 10, 8, 5 </p><p>(ii) 2, 2, 5, 5, 11, 11 </p><p>Median </p><p>4 numbers 4 numbers </p><p>Median = </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>4. Find the mean, mode and the median of the following data </p><p>Time ( hours ) 12 13 14 15 16 </p><p>Number of cars 3 5 10 6 6 </p><p>5. The following frequency distribution table shows the score of a group of students in a </p><p>quiz. Find the mean , mode and median </p><p>Score 5 6 7 8 9 10 </p><p>Number of students 5 6 4 3 8 4 </p><p>7.1.4 Determine modal class of grouped data from the frequency distribution table. </p><p>Modal class of a set of data is the value of class which occurs most frequently. The value of mode </p><p>can be obtained by drawing the histogram </p><p>Example 1: </p><p>The following table below shows the mark for 50 students in their Additional Mathematics test. </p><p>Find the modal class for the students. </p><p>Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 </p><p>Number of students 3 6 13 10 7 7 4 </p><p>7.1.5 Find mode from a histogram </p><p>Note: In drawing a histogram, class boundaries are used. </p><p>HISTOGRAM </p><p>Plot: frequency against boundary of the class </p><p>Frequency </p><p>Boundary of the class </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Example 1: </p><p>The following table below shows the mark for 50 students in their Additional Mathematics test. . </p><p>Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 </p><p>Number of students 3 6 13 10 7 7 4 </p><p>Table 3 </p><p>Complete the following table. </p><p>Mark Frequency , f Lower class boundary Upper class boundary </p><p>10 19 3 </p><p>From the table, draw a histogram, hence determine the mode from the histogram. </p><p>Calculate mean of grouped data </p><p>f</p><p>fxx where f is the frequency for each class </p><p> x is the corresponding class midpoint </p><p> Note : </p><p> Class mid point = 2</p><p>limitupper limit lower </p><p>Example 1: </p><p>1. The table below shows the marks obtained by 30 students in a Mathematics test. </p><p>Marks 30 - 39 40 - 49 50 - 59 60 -69 </p><p>Number of students 4 8 12 6 </p><p>Find the mean for the marks. </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Solution : </p><p>7.1.7 Calculating median of grouped data from the cumulative frequency distribution table </p><p>Median, m = L + Cf</p><p>FN</p><p>m</p><p>2</p><p>1</p><p> , where </p><p> Note: Size of class interval = upper class boundary lower class boundary </p><p>Activity : </p><p>The following table shows the mark for 40 students in their Additional Mathematics test. Find </p><p>the median by using the formula </p><p>Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 </p><p>Number of students 3 5 7 9 10 4 2 </p><p>TABLE 1 </p><p>Marks Number of students, f Class mark ,x fx </p><p> f fx </p><p>L = Lower boundary of the class in which the median </p><p>lies. </p><p>N = total frequency </p><p>C = Size of class interval </p><p>F = cumulative frequency before the class in which the </p><p>median lies. </p><p>fm = frequency of the class in which the median lies </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Solution: </p><p>Mark Frequency , f Cumulative </p><p>frequency, F </p><p>10 19 3 </p><p>7.1.8 Estimate median of grouped data from an ogive </p><p>OGIVE </p><p>Plot: cumulative frequency against upper class boundary </p><p>Cumulative </p><p>frequency </p><p>Upper class boundary </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Refer to table 1 </p><p>Construct a cumulative frequency table and then draw cumulative frequency curve (ogive) </p><p>From the graph, find the median weight. </p><p>Mark Frequency , f Upper class </p><p>Boundary </p><p>Cumulative </p><p>frequency, F </p><p>Activity 2: </p><p>1. The following table shows the marks obtained by 30 students in Mathematics test. Find the mean for the data. </p><p>Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 </p><p>Number of students 3 6 13 10 7 7 4 </p><p>Solution : </p><p>Mark Number of students, f Class mark ,x fx </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>2. The data below shows the scores obtained by 30 students in a game. </p><p>0 3 3 6 8 9 9 10 10 11 </p><p>5 0 7 6 9 10 12 17 5 4 </p><p>2 5 8 8 10 11 7 8 12 11 </p><p>Using 3 scores as a size of class interval, construct a frequency distribution table for its data. </p><p>Then find </p><p>(a) the mean (b) median (c) modal class </p><p> Note : Students have to draw the frequency distribution table as below. </p><p>Number of scores x f fx Cumulative </p><p>frequency </p><p>3. The table below shows the age of factory workers in 1995. </p><p>Age(year) 20-25 26-31 32-37 38-43 44-49 50-55 56-61 </p><p>Number of workers 5 24 16 20 13 12 10 </p><p>(i) Draw a histogram and hence estimate the mode of the data. (ii) Find the median without using an ogive. </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>7.1.9 Determine the effects on mode, median and mean for a set of data when </p><p>(a) Every value of the data is change uniformly </p><p>Measures of </p><p>central </p><p>tendency </p><p>Added by k Subtracted by k Multiplied by k Divided by k </p><p>New mean Original mean + k Original mean - k k(Original mean) </p><p>k</p><p>mean original </p><p>New mode Original mode + k Original mode - k K(Original mode) </p><p>k</p><p>mode original </p><p>New median Original median+ k Original median - k K(Original median) </p><p>k</p><p>median original </p><p>1. The mean, mode and median of a set data are 7.4, 9 and 8 respectively. Find the new </p><p>mean, mode and median if every value of the set data is </p><p>a) divided by 2 b) subtracted by 4 c) multiplied by 3 d) added by 3 </p><p>2. Find the mean of 12, 14, 16, 18, 20. By using the result, find the mean of 8 ,10 ,12 ,14,16 </p><p>7.1 MEASURES OF DISPERSION </p><p>7.2.1 Finding the measures of dispersion of ungrouped data </p><p>Formulae: </p><p>(i) Range of ungrouped data = largest value smallest value (ii) Interquartile range = Upper quartile lower quartile </p><p> = 3Q 1Q </p><p>(iii) Variance, 2 : </p><p>2</p><p>2)(</p><p>N</p><p>xxi 2</p><p>2</p><p>2 xN</p><p>xor </p><p> , where </p><p>N</p><p>xx</p><p>(iv) Standard deviation, : = iancevar </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Example : </p><p>For the given ungrouped set of data, find </p><p>(i) range (ii) interquartile range (iii) variance (iv) standard deviation </p><p>(a) 3 , 6 , 8 , 12 , 15 , 16 </p><p>(b) 3 , 5 , 6 , 7 , 8 , 4 , 8 , 9 , 10 , 12 , 14 </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Activity 3: </p><p>1. Find range, interquartile range, variance and standard deviation for each set of the </p><p>following data. </p><p>(a) 10 , 7 , 19 , 13 , 14 , 10 </p><p>(b) 4 , 3 , 2 , 7 , 9 , 10 , 12 , 6 , 15 </p><p>(d) 4 , 12 , 15 , 10 , 7 , 6 , 1 </p><p>2. Given that the mean of set data 5 , 7 , x , 11 , 12 is 9. </p><p>a. Find the value of x. b. Find the variance and the standard deviation of the data. </p><p>7.2.2 Finding measures of dispersion of grouped data </p><p>Example: </p><p>For the following table, </p><p>Find (i) range </p><p>(ii) interquartile range using ogive </p><p>Age ( year ) Number of workers Upper class boundary Cumulative frequency </p><p>25 29 16 </p><p>30 34 20 </p><p>35 39 24 </p><p>40 44 20 </p><p>45 49 14 </p><p>50 54 6 </p><p>Solution: </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>7.2.3 Determining the variance and standard deviation of grouped data. </p><p>Formulae : </p><p>N</p><p>xxfiance</p><p>2</p><p>2)(</p><p>var or 2 = 22</p><p>xf</p><p>fx</p><p> where </p><p>f</p><p>fxmeanx </p><p> standard deviation = 222)(</p><p>var xf</p><p>fxor</p><p>N</p><p>xxfiance </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Example 1: </p><p>From the following table, calculate the variance and standard deviation. </p><p>Mass ( kg ) Number of </p><p>students </p><p>46 50 3 </p><p>51 55 7 </p><p>56 60 10 </p><p>61 65 5 </p><p>66 70 6 </p><p>71-75 9 </p><p>Solution: </p><p>Example 2: </p><p>The heights of 100 men to the nearest cm are recorded as follows. Calculate the mean ( giving </p><p>your answer to one decimal place). Calculate the standard deviation for the height of the men. </p><p>Height , x(cm) Frequency, f </p><p>159 - 160 2 </p><p>161 162 14 </p><p>163 164 28 </p><p>165 166 26 </p><p>167 168 19 </p><p>169 170 9 </p><p>171 - 172 2 </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>Solution: </p><p>Class f x fx 2x fx2 </p><p>Activity 4: </p><p>1. The following table shows the number of durians sold in 50 consecutive days. </p><p>Number of durians Number of days </p><p>0 8 10 </p><p>9 17 15 </p><p>18 26 20 </p><p>27 - 35 5 </p><p>Calculate the variance and standard deviation </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>2. The table below shows the distribution of marks of 120 students in a Physics test. </p><p>Marks Number of </p><p>students </p><p>20 29 2 </p><p>30 39 14 </p><p>40 49 35 </p><p>50 59 50 </p><p>60 69 17 </p><p>70 - 79 2 </p><p> Calculate </p><p>a. mean b. median c. standard deviation </p><p> for this distribution. </p><p>7.2.4 Determining the effects on measures of dispersion when some values in a set of data are changed. </p><p>A. If every value of the data is changed uniformly </p><p>Measures Added by k Subtracted by k Multiplied by k Divided by k </p><p>New </p><p>Variance, 2 </p><p>2 </p><p>(original </p><p>variance) </p><p>2 </p><p>(original variance) )( 22 k </p><p>2</p><p>2</p><p>k</p><p>New </p><p>Standard </p><p>Deviation, </p><p>(original </p><p>standard </p><p>deviation) </p><p>(original standard </p><p>deviation) </p><p>)(k k</p><p>original </p><p>New Range </p><p> Original range Original range </p><p>range) original(k</p><p> k</p><p>range original </p><p>New </p><p>Interquartile </p><p>range </p><p>Original </p><p>interquartile </p><p>range </p><p>Original </p><p>interquartile range </p><p>k(original </p><p>interquartile </p><p>range) k</p><p>ileinterquart original</p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>B. If there are extreme values in the set of data </p><p>Range </p><p>Extreme values in a set of data will significantly increase the range of the set of data. </p><p>Interquartile range </p><p>Extreme values in a set of data will have little or no effect on the interquartile range. </p><p>Variance and Standard Deviation </p><p>Extreme values also significantly increase the value of standard deviation and variance but </p><p>standard deviation is affected to a smaller degree as compare to variance. </p><p>C. If certain values are added or removed </p><p>When a value is added or removed from a set of data, the effect on the measures of dispersion is </p><p>uncertain. In general, the range and the interquartile range are less affected as compared to the </p><p>variance and the standard deviation. Variance and standard deviation are more significantly </p><p>affected when the added or removed value has a greater difference from the mean. </p><p>Activity: </p><p>(a) The interquartile range and the standard deviation of a set of data are 5 and 2.5. Find the new </p><p>interquartile range and standard deviation if every value of the data is divided by 2 followed </p><p>by an addition of 10. </p><p>(b) Given a set of data 4, 5, 8, 12, 14, 18, 20. Determine the standard deviation of the set of data. </p><p>Explain how the standard deviation will change if a value of 100 is added to the set of data. </p><p>Activity 5: </p><p>1 The following table shows the results in five Additional Mathematics Test obtained by two </p><p>students A and B. </p><p>Students A 71 76 80 83 90 </p><p>Students B 40 67 95 98 100 </p><p> (a) Find the mean and the standard deviation of the result for each students. </p><p>(b) If a student with a more consistent performance is selected, which students would </p><p>be selected? Explain the reason for your selection. </p><p>2 Find the mean and standard deviation for the following set of numbers: 2, 3, 5, 8, 10. </p><p>Hence, using this result find the mean and standard deviation for </p><p>(a) five numbers : 7, 8, 10, 13, 15 </p><p>(b) five numbers : 10, 15, 25, 40, 50 </p><p>(c ) ten numbers : 2, 2, 3, 3, 5, 5, 8, 8, 10, 10 </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>3 The mean for 7, 14, 2x, x, 16, 9, 10 and 2 is 8. Find the value of </p><p>(a) x </p><p>(b) the standard deviation </p><p>(c ) standard deviation if each of the number is added by 2. </p><p>Enrichment </p><p>SPM 2006 </p><p>1. A set of positive integers consists of 2 , 5 and m. The variance for this set of integers is 14. Find the value of m. [ 3 marks ] </p><p>2. Table 1 below shows the frequency distribution of the scores of a group of pupils in a game. </p><p>Score Number of pupils </p><p>10 19 1 </p><p>20 29 2 </p><p>30 39 8 </p><p>40 49 12 </p><p>50 59 k </p><p>60 69 1 </p><p> Table 1 </p><p> a. It is given that the median score of the distribution is 42. Calculate the value of k. </p><p> [ 3 marks ] </p><p> b. Use the graph paper to answer this question </p><p> Using a scale of 2 cm to 10 scores on the horizontal axis and 2 cm to 2 pupils on the </p><p> vertical axis, draw a histogram to represent the frequency distribution of the scores. </p><p> Find the mode score [ 4 marks ] </p><p> c. What is the mode score if the score of each pupil is increased by 5? [ 1 mark ] </p><p>SPM 2005 </p><p>3. The mean of four numbers is m .The sum of the squares of the numbers is 100 and the </p><p>standard deviation is 3k. Express m in terms of k. [ 3 marks ] </p><p>4. Diagram 2 is a histogram which represents the distribution of the marks obtained by 40 pupils in a test. </p></li><li><p>Statistics </p><p>zefry@sas.edu.my </p><p>(a) Without using an ogive, calculate the median mark. [ 3 marks ] </p><p>(b) Calculate the standard deviation of the distribution. [ 4 marks ] </p><p>SPM 2004 </p><p>5. A set of data consist of 10 numbers. The sum of the numbers is 150 and the sum of the squares of the numbers is 2 472. </p><p>(a) Find the mean and variance of the 10 numbers. </p><p>(b) Another number is added to the set of data and the mean is increased by 1. </p><p> Find </p><p>(i) the value of this number, (ii) the standard deviation of the set of 11 numbers. [ 4 marks ] </p><p>SPM 2003 </p><p>6. A set of examination marks x1, x2, x3, x4, x5, x6 has a mean of 5 and a standard deviation of 1.5. </p><p>(a)...</p></li></ul>