INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables You have chosen to study: Please choose a question to attempt from the following: EXIT Back to Unit 2 Menu 56 Stem & Leaf Dot Plot Cum Freq Table Dot to boxplot Stem to boxplot Piechart

(a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory … n = = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Reveal answer EXIT Get hint

(a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory … n = = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end What would you like to do now?

(a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory … n = = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments EXIT Graphs etc Menu median = £183 Q1 = £171 Q3 = £195 = £12 = 2/5 What would you like to do now?

Comments Begin Solution Continue Solution Question 1 Menu Back to Home n = = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £ There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 position = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 What would you like to do now? (NOT 3!!!)

Comments Begin Solution Continue Solution Question 1 Menu Back to Home n = = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 4. Use SIQR = ½ (Q 3 – Q 1 ) / 2 (iii) SIQR = ½(Q3 – Q1) = (£195 - £171)  2 = £12

Question 1 5. Use P = no of favourable / no of data n = = £174 (b)What is the probability that someone chosen at random earns less than £180? No of favourable ( under £180) = 10 No of data = n = 25 (b) Prob(under £180) = 10 / 25 = 2 / 5. Comments Begin Solution Continue Solution Menu Back to Home

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £ There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 Median: the middle number in the ordered list. 25 numbers in the list. 1 – numbers on either side of the medianmedian is the 13th number in order.

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £ There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q1Q1 12 numbers before median. 6 numbers either side of Q 1 is midway between the 6th and 7th number.

Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £ There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q3Q3 12 numbers after median. 6 numbers either side of Q 3 is midway between the 19th and 20th number.

Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Go to full solution Go to Comments Reveal answer Get hint EXIT

Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Establish lowest & highest values and draw line with scale. Plot a dot for each piece of data and label diagram. For probability use: P = no of favourable / no of data Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT What would you like to do now?

Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Tightly clustered 3/10 EXIT Go to full solution Go to Comments Graphs etc Menu CLICK

Question 2 1. Establish lowest & highest values and draw line with scale Illustrate this using a dot plot (a) Lowest = 28 & highest = 31. Weights in g 2. Plot a dot for each piece of data and label diagram. Comments Begin Solution Continue Solution Menu Back to Home

Question 2 3. Make sure you know the possible descriptions of data Weights in g What type of distribution does this show? (b) Tightly clustered distribution. Comments Begin Solution Continue Solution Menu Back to Home

Question Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. If a bag is chosen at random what is the probability it will be heavier than the modal weight? Mode! Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 3. Make sure you know the possible descriptions of data Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

Comments 3. Make sure you know the possible descriptions of data Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

Comments 3. Make sure you know the possible descriptions of data Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

Comments Probability = Number of favourable outcomes Number of possible outcomes Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. Mode! To calculate simple probabilities: Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 3 The results for a class test were (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? Go to full solution Go to Comments Reveal answer Get hint EXIT Graphs etc Menu

Charts, Graphs & Tables : Question 3 The results for a class test were (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? Establish lowest & highest values and draw table. Complete each row 1 step at a time, calculating running total as you go. For probability use: P = no of favourable / no of data Use median = (n+1) / 2 to establish in which row median lies. EXIT Go to full solution Go to Comments Graphs etc Menu Reveal answer What would you like to do now?

Charts, Graphs & Tables : Question 3 The results for a class test were (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? CLICK Median = 14 1/3 EXIT Go to full solution Go to Comments Graphs etc Menu

Question 3 1. Establish lowest & highest values and draw a table. (a) Lowest = 10 & highest = (a)Construct a cumulative frequency table for this data. Mark Frequency Cum Frequency Complete each row 1 step at a time, calculating running total as you go. Comments Begin Solution Continue Solution Menu Back to Home

Question 3 3. Use median = (n+1) / 2 to establish in which row median lies Mark Frequency Cum Frequency (b) What is the median for this data? For 30 values median is between 15th & 16th both of which are in row 14. Median Mark = 14 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Question Mark Frequency Cum Frequency (c) What is the probability that a pupil selected at random scored under 14? No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / Use P = no of favourable / no of data Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments For 30 values median is between 15th & 16th both of which are in row 14. Median = 14 Mark Freq Cum Freq Median: 1 – 15 Q Find the mark at which the cumulative frequency first reaches between 15 th and 16 th number. Median = 14 Menu Back to Home Next Comment

Comments Mark Freq Cum Freq No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / 3. Probability = Number of favourable outcomes Number of possible outcomes To calculate simple probabilities: Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end remember bigger SIQR means more variation (spread) in data. EXIT Go to full solution Go to CommentsReveal answer Graphs etc Menu What would you like to do now?

Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Median = 50 So Q1 = 49 So Q3 = 52 the data is distributed more widely than (or not as clustered as) the above data EXIT Menu Full solutionComments CLICK

Question (a)Find the (i) median (ii) lower quartile (iii) upper quartile (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 Comments Begin Solution Continue Solution Menu Back to Home

Question (b) Construct a boxplot using this data. 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q (b)Lowest = 48, Q1 = 49, Q2 = 50, Q3 = 52 & Highest = 58. Comments Begin Solution Continue Solution Menu Back to Home

Question Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c) In a second sample the semi-interquartile range was 2.5. How does this compare? (c)For above sample SIQR = ( )  2 = 1.5 In a sample where the SIQR is 2.5 the data is distributed more widely than (or not as clustered as) the above data Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = numbers in the list: Q2Q2 11 numbers on either side of the median The median: Menu Back to Home Next Comment

Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 For quartiles: Q2Q2 Q1Q1 Q3Q Q2Q2 Now count through the list until you reach the 6 th, 12 th,and 18 th number in the list. Menu Back to Home Next Comment

Comments 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c)For above sample SIQR = ( )  2 = 1.5 In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data The semi-interquartile range is a measure of the range of the “middle” 50%. S.I.R. = (Q 3 - Q 1 ) 1212 Remember: when asked to compare data always consider average and spread. It is a measure of how spread-out and so how “consistent” or “reliable” the data is. Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results = 114kg Full solution Comments Reveal answer Get hint EXIT

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results = 114kg Use median position = (n+1) / 2 to find median position Q1 is midpoint from start to median Q3 is midpoint from median to end When comparing two data sets comment on spread and average Full solution Comments Reveal answer Menu EXIT What now?

Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results = 114kg median = 87 Q1 = 77 Q3 = 99 Full solution Comments Menu EXIT CLICK

Question 5 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 26 then the median is between 13 th & 14 th value ie median = 87 (ii) so Q1 = There are 13 values before median so Q 1 position is 6 th value 3. There are 13 values after median so Q 3 position is 20 th position so Q3 = = 114kg (a)Find the median, lower & upper quartiles for this data. Comments Begin Solution Continue Solution Menu Back to Home

Question = 114kg 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = (b) Use the data to construct a boxplot. Comments Begin Solution Continue Solution Menu Back to Home

Question 5 5. Compare spread and relevant average. (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results (c)Lightest has put on weight – lowest now 65, heaviest 3 have lost weight – highest now 115, median same but overall spread of weights has decreased as Q3-Q1 was 22 but is now only 15. Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

Comments 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = Box Plot : LowestHighestQ1Q1 Q2Q2 Q3Q3 Remember: To draw a boxplot you need a “five-figure summary”: five-figure summary Menu Back to Home Next Comment

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? angle 360° = amount 630 EXIT Go to full solution Go to Comments Reveal answer Graphs etc Menu What would you like to do now?

Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? = 252 = 48° EXIT Go to full solution Go to Comments Graphs etc Menu What would you like to do now?

Question 6 clubbin g 144° x°x° theatre cinemaWatching TV How many people went clubbing? 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount ° 360° = amount x amount = 144 x 630 amount = 144 x 630  360 = 252 Comments Begin Solution Continue Solution Menu Back to Home

Question 6 clubbin g 144° x°x° theatre cinemaWatching TV 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = x angle = 360° x 84 angle = 360° x 84  630 = 48° (b) If 84 people went to the theatre then how big is x°? Comments Begin Solution Continue Solution Menu Back to Home

Comments 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount ° 360° = amount x amount = 144 x 630 amount = 144 x 630  360 = 252 Can also be tackled by using proportion: Amount = x Menu Back to Home Next Comment

Comments Can also be tackled by using proportion: 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = x angle = 360° x 84 angle = 360° x 84  630 = 48° 84 = x 630 x x = 84 x 360 x = x Menu Back to Home Next Comment End of graphs, charts etc.