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Adding all the frequencies therefore gives the total number of items represented by the table."},{"text":"Add the values, not the frequencies.","correct":false,"explanation":"Adding the listed values ignores how many times each one occurs. The total number of items is found by adding the frequencies, not the values."},{"text":"Multiply the frequencies, not the values.","correct":false,"explanation":"Multiplying frequencies does not count the observations and can produce a very large, meaningless result. Add the frequencies to obtain the total number of items."},{"text":"Add the value-frequency products, not the values.","correct":false,"explanation":"Adding value-frequency products finds the total of all data values, which is used for the mean. The number of items is found by adding the frequencies."}]},{"id":"41ce2c93-6498-a0b5-2e53-16c6582fd89c","cardType":"basic","points":1,"front":"Points frequencies are 3, 5, 4, 2. How many games were played?","back":"14 games.","mcqOptions":[{"text":"14 games.","correct":true,"explanation":"The frequencies count the games in each points category, so the total is $3+5+4+2=14$. Therefore, 14 games were played."},{"text":"12 games.","correct":false,"explanation":"This total misses one of the frequency groups. All four frequencies must be added: $3+5+4+2=14$, so 14 games were played."},{"text":"10 games.","correct":false,"explanation":"This comes from adding only the first three frequencies and omitting the final frequency of 2. Including every row gives $3+5+4+2=14$ games."},{"text":"16 games.","correct":false,"explanation":"This likely double-counts the final frequency of 2. Each frequency is included once, giving $3+5+4+2=14$ games in total."}]},{"id":"fc74617d-52b1-8d2d-b790-7dcfa2f37c53","cardType":"cloze","points":null,"front":"How many times a value, score, or group occurs is its [ ].","back":"How many times a value, score, or group occurs is its frequency.","mcqOptions":[{"text":"frequency","correct":true,"explanation":"Frequency is the number of times a particular value, score, or group appears in a data set. It is the count recorded for that category."},{"text":"range","correct":false,"explanation":"The range measures the spread of data by subtracting the lowest value from the highest. The number of times a value occurs is its frequency."},{"text":"median","correct":false,"explanation":"The median is the middle value after the data are ordered. The count of how many times a value or group occurs is its frequency."},{"text":"interval","correct":false,"explanation":"An interval is a span of values used to group data. The number of observations within a value or group is called its frequency."}]},{"id":"65293c2b-4245-526b-bfe9-9788d6590623","cardType":"basic","points":1,"front":"For the mean from an exact frequency table, what is multiplied before adding?","back":"Multiply each value by its frequency.","mcqOptions":[{"text":"Multiply each value by its frequency.","correct":true,"explanation":"Multiplying each value by its frequency gives the total contribution from every occurrence of that value. These products are then added before dividing by total frequency."},{"text":"Multiply each value by the total frequency.","correct":false,"explanation":"Using the total frequency for every value treats each score as if it occurred equally often. Each value must instead be multiplied by its own frequency."},{"text":"Multiply each frequency by itself.","correct":false,"explanation":"Squaring frequencies does not include the data values and therefore cannot produce the data total. Multiply each value by its corresponding frequency before adding."},{"text":"Multiply each value by the next frequency.","correct":false,"explanation":"Pairing a value with the next row’s frequency mismatches the data. Each value must be multiplied by the frequency recorded in the same row."}]},{"id":"283e0ed7-541b-4d28-0173-c56fcf2e9a96","cardType":"cloze","points":null,"front":"For exact frequency data, mean = total of [ ] divided by total [ ].","back":"For exact frequency data, mean = total of value-frequency products divided by total frequency.","mcqOptions":[{"text":"For exact frequency data, mean = total of value-frequency products divided by total frequency.","correct":true,"explanation":"Value-frequency products give the total of all observations, while total frequency gives the number of observations. Dividing the first total by the second gives the mean."},{"text":"For exact frequency data, mean = total of values divided by total frequency.","correct":false,"explanation":"Adding each listed value once ignores repeated observations. The numerator must be the total of value-frequency products, divided by the total frequency."},{"text":"For exact frequency data, mean = total of value-frequency products divided by total number of rows.","correct":false,"explanation":"The number of rows counts distinct values, not observations. The total of value-frequency products must be divided by total frequency to calculate the mean."},{"text":"For exact frequency data, mean = total of frequencies divided by total values.","correct":false,"explanation":"This reverses and omits the required quantities. The mean is the total of value-frequency products divided by the total frequency."}]},{"id":"746ce451-e88c-90b7-e980-97b8ccf6c1d8","cardType":"basic","points":2,"front":"Why is adding the different scores and dividing by the number of rows wrong?","back":"Frequencies show some scores occurred more often.","mcqOptions":[{"text":"Frequencies show some scores occurred more often.","correct":true,"explanation":"Different frequencies mean that some scores appear more times than others. The mean must account for this by weighting each score using its frequency."},{"text":"Values show some scores are larger than others.","correct":false,"explanation":"Differences between score sizes do not explain the error. Adding rows once ignores repetition; frequencies show how often each score occurred and must be used as weights."},{"text":"Rows each represent the same number of results.","correct":false,"explanation":"Rows do not necessarily represent equal numbers of results because their frequencies may differ. The frequencies show that some scores occurred more often than others."},{"text":"Products make every score occur exactly once.","correct":false,"explanation":"Value-frequency products account for repeated scores rather than making each occur once. Frequencies show how often scores occurred, so simply averaging the listed rows is wrong."}]},{"id":"2b2b90d6-16ed-5367-e75e-8480902a1bdc","cardType":"basic","points":1,"front":"If the product total is 49 and total frequency is 27, what is the mean to 1 decimal place?","back":"1.8","mcqOptions":[{"text":"1.8","correct":true,"explanation":"The mean is $49\\div27=1.814\\ldots$. The hundredths digit is 1, so rounding to one decimal place gives 1.8."},{"text":"2.0","correct":false,"explanation":"This rounds $49\\div27=1.814\\ldots$ to the nearest whole number before writing a decimal place. Rounding directly to one decimal place gives 1.8."},{"text":"0.6","correct":false,"explanation":"This results from reversing the division and calculating $27\\div49$. For a frequency-table mean, divide the product total by total frequency: $49\\div27=1.8$ to one decimal place."},{"text":"1.9","correct":false,"explanation":"The value $49\\div27$ is $1.814\\ldots$, and the hundredths digit is 1, not 5 or more. It therefore rounds down to 1.8."}]},{"id":"6c534391-ea6f-1b28-302b-3a491f1c6cca","cardType":"cloze","points":null,"front":"A row with frequency 0 contributes [ ] to the total.","back":"A row with frequency 0 contributes nothing to the total.","mcqOptions":[{"text":"nothing","correct":true,"explanation":"A frequency of 0 means that the value occurs no times. Its value-frequency product is therefore 0, so the row adds nothing to the total."},{"text":"its value","correct":false,"explanation":"The value is not included once when its frequency is 0. Multiplying the value by 0 gives 0, so the row contributes nothing."},{"text":"its frequency","correct":false,"explanation":"The frequency is used as a multiplier, not added as the row’s contribution. Since the frequency is 0, the value-frequency product contributes nothing."},{"text":"one item","correct":false,"explanation":"A frequency of 0 means there are no items in that row, not one item. Its value-frequency product is 0, so it contributes nothing."}]},{"id":"254b799c-a2c4-ba6e-1c4b-c95de9b539e5","cardType":"basic","points":1,"front":"If 2 goals were scored in $x$ matches, what is that row’s goal total?","back":"$2x$ goals.","mcqOptions":[{"text":"$2x$ goals.","correct":true,"explanation":"Each of the $x$ matches contributes 2 goals, so multiplying frequency by goals per match gives a row total of $2 \\times x=2x$ goals."},{"text":"$2+x$ goals.","correct":false,"explanation":"Adding the number of goals to the number of matches confuses repeated values with a sum. Two goals in each of $x$ matches gives $2 \\times x=2x$ goals."},{"text":"$\\frac{x}{2}$ goals.","correct":false,"explanation":"Dividing by 2 would halve the number of matches rather than total the goals. The row total is frequency multiplied by value, so it is $2x$ goals."},{"text":"$x^2$ goals.","correct":false,"explanation":"Squaring the number of matches does not use the 2 goals scored per match correctly. Multiplying 2 by the frequency $x$ gives $2x$ goals."}]},{"id":"b6f99c7b-185a-66d4-41c9-685ddbbea3d8","cardType":"basic","points":1,"front":"Known goal total is 23, total goals are 39, and the unknown row contributes $2x$. What equation finds $x$?","back":"$23+2x=39$","mcqOptions":[{"text":"$23+2x=39$","correct":true,"explanation":"The known rows contribute 23 goals and the unknown row contributes $2x$ goals. Their sum must equal the overall total of 39 goals."},{"text":"$23+x=39$","correct":false,"explanation":"This treats $x$ as the row’s goal total, forgetting that each of those matches contributes 2 goals. The unknown contribution is $2x$, giving $23+2x=39$."},{"text":"$23+2x=62$","correct":false,"explanation":"Adding 23 and 39 to form the right-hand side double-counts the known contribution. The known and unknown contributions together equal 39, so $23+2x=39$."},{"text":"$39+2x=23$","correct":false,"explanation":"This reverses the total and the known contribution. The smaller known total of 23 is increased by $2x$ to reach 39, so $23+2x=39$."}]},{"id":"cf90c638-96f9-af95-8c45-1445a62f13f3","cardType":"basic","points":1,"front":"Solve $23+2x=39$.","back":"$x=8$","mcqOptions":[{"text":"$x=8$","correct":true,"explanation":"Subtracting 23 from both sides gives $2x=16$. Dividing both sides by 2 then gives $x=8$."},{"text":"$x=16$","correct":false,"explanation":"This stops after subtracting 23, when the equation is $2x=16$. Dividing by the coefficient 2 is still required, giving $x=8$."},{"text":"$x=7$","correct":false,"explanation":"This comes from an incorrect subtraction of 23 from 39. The correct difference is 16, so $2x=16$ and therefore $x=8$."},{"text":"$x=31$","correct":false,"explanation":"This does not isolate the $2x$ term correctly. Subtract 23 from 39 to obtain 16, then divide by 2 to get $x=8$."}]},{"id":"93b4f6dc-3f07-e5ca-585c-f2ecc06cae73","cardType":"cloze","points":null,"front":"The mode is the value with the [ ].","back":"The mode is the value with the highest frequency.","mcqOptions":[{"text":"highest frequency","correct":true,"explanation":"The mode is the data value that occurs most often, so it is identified by the highest frequency in a frequency table."},{"text":"lowest frequency","correct":false,"explanation":"The lowest frequency identifies the value occurring least often, not the mode. The mode is the value with the highest frequency."},{"text":"greatest value","correct":false,"explanation":"The largest data value is not necessarily the one occurring most often. The mode depends on frequency, so it is the value with the highest frequency."},{"text":"highest cumulative frequency","correct":false,"explanation":"Cumulative frequency increases towards the total and does not identify the most common value directly. The mode is found using the highest ordinary frequency."}]},{"id":"bcf6cdc3-3039-d6e9-39a5-ef48e46d2337","cardType":"basic","points":1,"front":"For 32 values, which positions locate the median?","back":"The 16th and 17th values.","mcqOptions":[{"text":"The 16th and 17th values.","correct":true,"explanation":"An even-sized set has two central positions. For 32 ordered values, these are positions $32\\div2=16$ and $16+1=17$."},{"text":"The 16th value.","correct":false,"explanation":"Using only the 16th value applies the odd-data rule incorrectly. Since 32 is even, the median is located using both the 16th and 17th values."},{"text":"The 17th value.","correct":false,"explanation":"Using only the 17th value ignores the other central position. For 32 ordered values, the middle lies between the 16th and 17th values."},{"text":"The 15th and 16th values.","correct":false,"explanation":"These positions are shifted one place too low. Half of 32 is 16, so the two central positions are the 16th and 17th values."}]},{"id":"aa3de2c7-1b07-80d6-5ade-1d4dad4c39b1","cardType":"basic","points":2,"front":"Why use cumulative frequency to find the median?","back":"It locates the middle value without writing the list.","mcqOptions":[{"text":"It locates the middle value without writing the list.","correct":true,"explanation":"Cumulative frequency shows which row contains the central position or positions, allowing the median to be found without expanding the frequency table into an ordered list."},{"text":"It locates the most frequent value without writing the list.","correct":false,"explanation":"The most frequent value is the mode and is found from the largest ordinary frequency. Cumulative frequency is used to locate the middle value for the median."},{"text":"It calculates the exact mean without writing the list.","correct":false,"explanation":"The mean requires the sum of the values and their frequencies, not cumulative frequency. Cumulative frequency instead locates the middle value needed for the median."},{"text":"It identifies the largest value without writing the list.","correct":false,"explanation":"The largest value can be read from the final non-zero row and does not require cumulative frequency. Cumulative totals are useful for locating the median’s middle position."}]},{"id":"40af13a9-6bb9-de8a-a874-8dd4ea6ef35c","cardType":"basic","points":1,"front":"Cumulative frequencies are up to 0 goals = 5 and up to 1 goal = 17. Where are the 16th and 17th values?","back":"In the 1-goal row.","mcqOptions":[{"text":"In the 1-goal row.","correct":true,"explanation":"Positions 1 to 5 are in the 0-goal row, while positions 6 to 17 are in the 1-goal row. Therefore, both central values are 1."},{"text":"In the 0-goal row.","correct":false,"explanation":"The 0-goal row contains only positions 1 to 5 because its cumulative frequency is 5. Positions 16 and 17 fall in the 1-goal row."},{"text":"In the 2-goal row.","correct":false,"explanation":"The cumulative frequency has already reached 17 by the end of the 1-goal row. Therefore, positions 16 and 17 occur before the 2-goal row."},{"text":"Across the 0-goal and 1-goal rows.","correct":false,"explanation":"The boundary between these rows is after position 5, not between positions 16 and 17. Both positions lie within the 1-goal row."}]},{"id":"e0bff81a-d0eb-6e0e-37a5-116ee2ba47f8","cardType":"cloze","points":null,"front":"For grouped data, the mean is only an [ ] because exact values are [ ].","back":"For grouped data, the mean is only an estimate because exact values are unknown.","mcqOptions":[{"text":"For grouped data, the mean is only an estimate because exact values are unknown.","correct":true,"explanation":"Grouped data gives intervals rather than individual observations, so calculations normally use class midpoints. Because the original exact values are unknown, the resulting mean is an estimate."},{"text":"For grouped data, the mean is only an exact value because exact values are unknown.","correct":false,"explanation":"Unknown individual values prevent an exact calculation, even though class midpoints can represent each interval. The resulting mean is therefore an estimate, not an exact value."},{"text":"For grouped data, the mean is only an estimate because exact values are known.","correct":false,"explanation":"If every exact value were known, the mean could be calculated exactly. Grouped tables provide only intervals, so exact values are unknown and the mean is an estimate."},{"text":"For grouped data, the mean is only an exact value because exact values are known.","correct":false,"explanation":"This describes ungrouped raw data rather than grouped data. In grouped data the individual values are unknown, so midpoints are used and the mean is only an estimate."}]},{"id":"4fa50128-6e0f-c27c-c1d4-4a102aa4ca91","cardType":"basic","points":1,"front":"How should you find each midpoint in grouped data?","back":"Use $\\frac{\\text{lower}+\\text{upper}}{2}$ for that interval.","mcqOptions":[{"text":"Use $\\frac{\\text{lower}+\\text{upper}}{2}$ for that interval.","correct":true,"explanation":"The midpoint is halfway between the interval boundaries, so add the lower and upper values and divide their sum by 2."},{"text":"Use $\\frac{\\text{upper}-\\text{lower}}{2}$ for that interval.","correct":false,"explanation":"Subtracting the boundaries and halving gives half the class width, not its midpoint. Add the lower and upper boundaries, then divide by 2."},{"text":"Use $\\frac{\\text{lower}+\\text{upper}}{\\text{upper}}$ for that interval.","correct":false,"explanation":"Dividing by the upper boundary does not locate the centre of an interval. The midpoint is the sum of both boundaries divided by 2."},{"text":"Use $\\frac{\\text{lower}+\\text{upper}}{\\text{lower}}$ for that interval.","correct":false,"explanation":"Dividing by the lower boundary produces a ratio rather than the interval’s central value. Add both boundaries and divide the result by 2."}]},{"id":"68729cce-0468-598b-f628-fbc7cb409de8","cardType":"basic","points":1,"front":"What is the midpoint of over 10, up to 15 minutes?","back":"12.5 minutes.","mcqOptions":[{"text":"12.5 minutes.","correct":true,"explanation":"The midpoint is halfway between 10 and 15, so $(10+15)\\div2=12.5$ minutes."},{"text":"2.5 minutes.","correct":false,"explanation":"This is half the class width, found from $(15-10)\\div2$, rather than the midpoint. The midpoint is $(10+15)\\div2=12.5$ minutes."},{"text":"10.5 minutes.","correct":false,"explanation":"Adding $0.5$ to the lower boundary does not find the centre of a five-minute interval. Averaging 10 and 15 gives 12.5 minutes."},{"text":"15.5 minutes.","correct":false,"explanation":"Adding $0.5$ to the upper boundary places the value outside the interval. The average of the two boundaries is 12.5 minutes."}]},{"id":"3e5390c5-3f36-7a6b-e358-ad4f5226a978","cardType":"cloze","points":null,"front":"For grouped data, estimated mean = total of [ ] divided by total [ ].","back":"For grouped data, estimated mean = total of midpoint-frequency products divided by total frequency.","mcqOptions":[{"text":"For grouped data, estimated mean = total of midpoint-frequency products divided by total frequency.","correct":true,"explanation":"Each class midpoint represents its observations, so multiplying each midpoint by its frequency estimates the total. Dividing the product total by total frequency gives the estimated mean."},{"text":"For grouped data, estimated mean = total of class widths divided by total frequency.","correct":false,"explanation":"Class widths describe interval sizes, not the estimated total of the data. Use the total of midpoint-frequency products divided by total frequency."},{"text":"For grouped data, estimated mean = total of midpoint-frequency products divided by total class width.","correct":false,"explanation":"Dividing by total class width ignores how many observations there are. The midpoint-frequency product total must be divided by total frequency."},{"text":"For grouped data, estimated mean = total of midpoints divided by total number of classes.","correct":false,"explanation":"Averaging the class midpoints gives every class equal importance, even when frequencies differ. Weight each midpoint by frequency, then divide by total frequency."}]},{"id":"40503cb0-2424-5b63-4a9c-7045df4f1b45","cardType":"basic","points":1,"front":"If the grouped-data product total is 430 and total frequency is 20, what is the estimated mean?","back":"21.5 minutes.","mcqOptions":[{"text":"21.5 minutes.","correct":true,"explanation":"The estimated mean is the midpoint-frequency product total divided by total frequency, so $430\\div20=21.5$ minutes."},{"text":"8.6 minutes.","correct":false,"explanation":"This results from dividing 430 by an incorrect frequency of 50. Using the stated total frequency gives $430\\div20=21.5$ minutes."},{"text":"21 minutes.","correct":false,"explanation":"Rounding before completing the calculation loses the required precision. The exact division is $430\\div20=21.5$ minutes."},{"text":"22.5 minutes.","correct":false,"explanation":"This comes from using 450 as the product total instead of 430. The stated values give $430\\div20=21.5$ minutes."}]},{"id":"183a693d-0c01-464c-e0cd-b76015529b10","cardType":"basic","points":1,"front":"Cumulative frequencies are up to 10 = 5, up to 20 = 14, up to 30 = 25. For 40 people, what is the median class?","back":"Over 20, up to 30 minutes.","mcqOptions":[{"text":"Over 20, up to 30 minutes.","correct":true,"explanation":"For 40 people, the median lies around the 20th observation. Cumulative frequency rises from 14 to 25 in this class, so it contains the median."},{"text":"Over 10, up to 20 minutes.","correct":false,"explanation":"This class contains observations 6 to 14, so it does not contain the 20th observation. The median class is over 20, up to 30 minutes."},{"text":"Over 30, up to 40 minutes.","correct":false,"explanation":"Choosing a class after cumulative frequency 25 places the median too late in the data. The 20th observation lies over 20, up to 30 minutes."},{"text":"Over 0, up to 10 minutes.","correct":false,"explanation":"This first class contains only observations 1 to 5, not the middle observation. The 20th observation lies over 20, up to 30 minutes."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Averages from Frequency Tables","title":"Averages from Frequency Tables Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"537dbf7e-ede9-44ed-b841-12fb6f5c4122","walled":"$L38"}],"$L39","$L3a"]