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This ensures that larger bar areas correspond to more observations."},{"text":"Frequency density.","correct":false,"explanation":"Frequency density determines the bar’s height, not its area. The area represents frequency because multiplying frequency density by class width gives the frequency."},{"text":"Class width.","correct":false,"explanation":"Class width is represented by the bar’s horizontal width, not its area. The area of the bar represents the frequency in that class interval."},{"text":"Class boundary.","correct":false,"explanation":"Class boundaries mark the endpoints of each interval and do not correspond to bar area. The area of a histogram bar represents the class frequency."}]},{"id":"eaf4013c-93e7-4d5f-ab99-c88c0444c35d","cardType":"cloze","points":null,"front":"A histogram is used for grouped [ ].","back":"A histogram is used for grouped continuous data.","mcqOptions":[{"text":"continuous data","correct":true,"explanation":"Histograms display grouped continuous data because values can occur anywhere within each class interval, and the bars therefore touch to show this continuity."},{"text":"discrete data","correct":false,"explanation":"Discrete data consists of separate countable values and is usually shown using a bar chart. A histogram is used for grouped continuous data."},{"text":"categorical data","correct":false,"explanation":"Categorical data is divided into named groups and is normally displayed with a bar chart. Histograms are designed for grouped continuous data."},{"text":"nominal data","correct":false,"explanation":"Nominal data consists of unordered categories rather than numerical intervals, so a bar chart is appropriate. A histogram is used for grouped continuous data."}]},{"id":"3f38470d-2351-48a5-9767-a4c4b2585fed","cardType":"basic","points":1,"front":"In the class interval $0 \\le t < 5$, where does a value of exactly 5 go?","back":"The next class.","mcqOptions":[{"text":"The next class.","correct":true,"explanation":"The inequality $t < 5$ excludes 5 from this interval. Therefore, exactly 5 belongs in the next class, whose lower boundary is 5."},{"text":"The previous class.","correct":false,"explanation":"Treating the upper boundary as included ignores the strict inequality $t < 5$. Exactly 5 is excluded and must be placed in the next class."},{"text":"The current class.","correct":false,"explanation":"The current class contains values from 0 up to, but not including, 5. Therefore, exactly 5 must be placed in the next class."},{"text":"The nearest class.","correct":false,"explanation":"Class placement is determined by the inequalities, not by which interval seems nearest. Since $t < 5$ excludes 5, it belongs in the next class."}]},{"id":"f44305b2-6dc6-4283-b77f-3800815a4de4","cardType":"basic","points":1,"front":"Find the class width of $7 \\le t < 15$.","back":"8.","mcqOptions":[{"text":"8.","correct":true,"explanation":"Class width is found by subtracting the lower boundary from the upper boundary. Here, $15 - 7 = 8$, so the class width is 8."},{"text":"7.","correct":false,"explanation":"Subtracting an extra 1 incorrectly treats class width as a count of selected integer positions. Using the boundaries gives $15 - 7 = 8$."},{"text":"9.","correct":false,"explanation":"Counting both endpoints produces an off-by-one error and ignores that 15 is excluded. Class width is the boundary difference, $15 - 7 = 8$."},{"text":"22.","correct":false,"explanation":"Adding the two boundaries does not calculate the size of the interval. Class width is found by subtraction, giving $15 - 7 = 8$."}]},{"id":"d226efb5-6cd7-4881-919a-ca2ed2d79c54","cardType":"cloze","points":null,"front":"Class width = [ ] - [ ].","back":"Class width = upper boundary - lower boundary.","mcqOptions":[{"text":"Class width = upper boundary - lower boundary.","correct":true,"explanation":"Class width measures the distance between the interval’s boundaries, so it is calculated by subtracting the lower boundary from the upper boundary."},{"text":"Class width = lower boundary - upper boundary.","correct":false,"explanation":"Reversing the subtraction usually gives a negative result, although class width must be positive. Subtract the lower boundary from the upper boundary instead."},{"text":"Class width = class midpoint - lower boundary.","correct":false,"explanation":"The distance from the lower boundary to the midpoint is only half the class width. The full width is upper boundary minus lower boundary."},{"text":"Class width = upper boundary - class midpoint.","correct":false,"explanation":"The distance from the midpoint to the upper boundary is only half the interval. Class width is calculated as upper boundary minus lower boundary."}]},{"id":"e4a9300f-9870-4c0b-a0f5-7dc875dea1dd","cardType":"basic","points":1,"front":"Find the class width of $14 \\le t < 24$.","back":"10.","mcqOptions":[{"text":"10.","correct":true,"explanation":"Subtracting the lower boundary from the upper boundary gives the class width: $24 - 14 = 10$. The excluded upper endpoint does not alter this distance."},{"text":"9.","correct":false,"explanation":"Subtracting an additional 1 creates an off-by-one error. Class width is the difference between the boundaries, so $24 - 14 = 10$."},{"text":"11.","correct":false,"explanation":"Counting both endpoints as separate integer values is not how interval width is calculated. Subtracting the boundaries gives $24 - 14 = 10$."},{"text":"38.","correct":false,"explanation":"Adding the boundaries finds neither the width nor the number of values in the class. The correct calculation is $24 - 14 = 10$."}]},{"id":"7ec52f2f-ee5e-44ca-b4a4-bc62def83c86","cardType":"basic","points":1,"front":"What is the height of a histogram bar called?","back":"Frequency density.","mcqOptions":[{"text":"Frequency density.","correct":true,"explanation":"The vertical height of each histogram bar is frequency density. Multiplying this height by the class width gives the frequency represented by the bar’s area."},{"text":"Frequency.","correct":false,"explanation":"Frequency is represented by the bar’s area, not necessarily its height, because histogram classes may have unequal widths. The height is called frequency density."},{"text":"Class width.","correct":false,"explanation":"Class width is shown horizontally along the axis and determines the bar’s width. The vertical height of a histogram bar is called frequency density."},{"text":"Class boundary.","correct":false,"explanation":"Class boundaries are the endpoints marked along the horizontal axis, not vertical measurements. The height of a histogram bar is called frequency density."}]},{"id":"cf3256e4-9dc0-4c14-befb-7cef926545ea","cardType":"cloze","points":null,"front":"Frequency density is often shortened to [ ].","back":"Frequency density is often shortened to f.d..","mcqOptions":[{"text":"f.d.","correct":true,"explanation":"The standard abbreviation for frequency density is f.d., using the initial letters of the two words. It commonly labels the vertical axis of a histogram."},{"text":"f.","correct":false,"explanation":"The abbreviation f. refers to frequency rather than frequency density. Frequency density is shortened to f.d. and represents the height of a histogram bar."},{"text":"c.w.","correct":false,"explanation":"The abbreviation c.w. stands for class width, which is shown horizontally. Frequency density, the height of a histogram bar, is shortened to f.d."},{"text":"c.f.","correct":false,"explanation":"The abbreviation c.f. stands for cumulative frequency, a running total used in cumulative frequency graphs. Frequency density is shortened to f.d."}]},{"id":"ff428e9f-208a-456e-a1d8-ba3251197e45","cardType":"basic","points":1,"front":"How do you calculate frequency density from frequency and class width?","back":"Frequency density $= \\frac{\\text{frequency}}{\\text{class width}}$.","mcqOptions":[{"text":"Frequency density $= \\frac{\\text{frequency}}{\\text{class width}}$.","correct":true,"explanation":"Dividing the frequency by the class width accounts for unequal class intervals. The resulting frequency density determines the height of the corresponding histogram bar."},{"text":"Frequency density $= \\frac{\\text{class width}}{\\text{frequency}}$.","correct":false,"explanation":"This reverses the required division. Frequency density is calculated by dividing frequency by class width, not class width by frequency."},{"text":"Frequency density $= \\text{frequency} \\times \\text{class width}$.","correct":false,"explanation":"Multiplication is used to recover frequency from frequency density and class width. To find frequency density, divide the frequency by the class width."},{"text":"Frequency density $= \\frac{\\text{frequency}}{\\text{class midpoint}}$.","correct":false,"explanation":"The class midpoint identifies the centre of an interval but is not used to calculate bar height. Frequency density is frequency divided by class width."}]},{"id":"7a8f28bf-ec84-4c69-a6b2-d49c16793df3","cardType":"basic","points":1,"front":"A class has frequency 18 and width 6. Find its frequency density.","back":"3.","mcqOptions":[{"text":"3.","correct":true,"explanation":"Frequency density equals frequency divided by class width, so $18 \\div 6 = 3$. This would also be the height of the histogram bar."},{"text":"12.","correct":false,"explanation":"This comes from subtracting the class width from the frequency. Frequency density requires division, giving $18 \\div 6 = 3$."},{"text":"24.","correct":false,"explanation":"This comes from adding frequency and class width, which is not the frequency density formula. Dividing gives $18 \\div 6 = 3$."},{"text":"108.","correct":false,"explanation":"This multiplies frequency by class width instead of dividing. Frequency density is calculated as $18 \\div 6$, which equals $3$."}]},{"id":"ceddbf62-44aa-4785-a19e-9980a4f11d99","cardType":"basic","points":2,"front":"For $2 \\le t < 8$ with frequency 18, what height should the histogram bar have?","back":"3.","mcqOptions":[{"text":"3.","correct":true,"explanation":"The class width is $8 - 2 = 6$, so the frequency density and bar height are $18 \\div 6 = 3$."},{"text":"6.","correct":false,"explanation":"This is the class width, found from $8 - 2$, rather than the bar height. The height is frequency density: $18 \\div 6 = 3$."},{"text":"18.","correct":false,"explanation":"This uses the frequency directly as the height, which only works when class widths are equal and scaled appropriately. Here the height is $18 \\div 6 = 3$."},{"text":"108.","correct":false,"explanation":"This multiplies the frequency by the class width. Histogram height is frequency density, so the correct calculation is $18 \\div 6 = 3$."}]},{"id":"d27fad20-02d9-4eb2-bd59-6829686249ce","cardType":"cloze","points":null,"front":"In a histogram, frequency is represented by bar [ ], not just height.","back":"In a histogram, frequency is represented by bar area, not just height.","mcqOptions":[{"text":"area","correct":true,"explanation":"A histogram represents frequency through bar area because area equals class width multiplied by frequency density. Bar height alone represents frequency density."},{"text":"height","correct":false,"explanation":"Height represents frequency density, not frequency itself. Frequency is represented by bar area because multiplying frequency density by class width gives the frequency."},{"text":"width","correct":false,"explanation":"Width represents the size of the class interval, not its frequency. Frequency is represented by bar area, combining class width with frequency density."},{"text":"perimeter","correct":false,"explanation":"Perimeter has no statistical meaning in a histogram. Frequency is represented by bar area, calculated from class width multiplied by frequency density."}]},{"id":"6c07bb60-a51a-49c4-8b80-ddb1c30002ea","cardType":"basic","points":1,"front":"How do you calculate frequency from class width and frequency density?","back":"Frequency $= \\text{class width} \\times \\text{frequency density}$.","mcqOptions":[{"text":"Frequency $= \\text{class width} \\times \\text{frequency density}$.","correct":true,"explanation":"A histogram bar’s area represents frequency. Since its width is the class width and its height is the frequency density, multiplying them gives the frequency."},{"text":"Frequency $= \\frac{\\text{class width}}{\\text{frequency density}}$.","correct":false,"explanation":"Dividing class width by frequency density does not give bar area. Frequency is found by multiplying class width by frequency density."},{"text":"Frequency $= \\frac{\\text{frequency density}}{\\text{class width}}$.","correct":false,"explanation":"This divides the bar height by its width rather than calculating its area. Frequency equals class width multiplied by frequency density."},{"text":"Frequency $= \\text{class width} + \\text{frequency density}$.","correct":false,"explanation":"Adding width and height does not calculate a histogram bar’s area. Frequency is found by multiplying class width by frequency density."}]},{"id":"eca1a7cb-e7d0-4ca7-8f3a-94f6dd64a152","cardType":"basic","points":1,"front":"A bar has width 8 and frequency density 1.75. Find the frequency.","back":"14.","mcqOptions":[{"text":"14.","correct":true,"explanation":"Frequency is class width multiplied by frequency density. Therefore, $8 \\times 1.75 = 14$, which is the area represented by the bar."},{"text":"4.57.","correct":false,"explanation":"This comes from dividing $8$ by $1.75$ rather than finding the bar’s area. Multiplying gives the frequency: $8 \\times 1.75 = 14$."},{"text":"9.75.","correct":false,"explanation":"This adds the width and frequency density, but frequency corresponds to bar area. The correct calculation is $8 \\times 1.75 = 14$."},{"text":"6.25.","correct":false,"explanation":"This subtracts the frequency density from the class width, which does not calculate frequency. Multiplying $8$ by $1.75$ gives $14$."}]},{"id":"a433d1ad-5ec2-42c3-a261-ce16405cef2d","cardType":"basic","points":2,"front":"For $8 \\le t < 16$ with frequency density 1.75, find the frequency.","back":"14.","mcqOptions":[{"text":"14.","correct":true,"explanation":"The class width is $16 - 8 = 8$. Multiplying this by the frequency density gives the frequency: $8 \\times 1.75 = 14$."},{"text":"8.","correct":false,"explanation":"This is only the class width, calculated as $16 - 8$. Frequency is width multiplied by frequency density, so $8 \\times 1.75 = 14$."},{"text":"9.75.","correct":false,"explanation":"This adds the class width and frequency density instead of finding the bar’s area. The frequency is $8 \\times 1.75 = 14$."},{"text":"28.","correct":false,"explanation":"This commonly results from treating $1.75$ as $3.5$, doubling the density. Using the given density gives $8 \\times 1.75 = 14$."}]},{"id":"54921194-6dcf-4ac6-89f4-e7b6f2cbb36a","cardType":"cloze","points":null,"front":"To find frequency density, divide [ ] by [ ].","back":"To find frequency density, divide frequency by class width.","mcqOptions":[{"text":"To find frequency density, divide frequency by class width.","correct":true,"explanation":"Frequency density is the frequency per unit of class width. Dividing frequency by class width gives the correct height for a histogram bar."},{"text":"To find frequency density, divide class width by frequency.","correct":false,"explanation":"This reverses the numerator and denominator. Frequency density measures frequency per unit width, so frequency must be divided by class width."},{"text":"To find frequency density, divide frequency by class midpoint.","correct":false,"explanation":"The midpoint locates the centre of a class but does not determine histogram height. Frequency density is frequency divided by class width."},{"text":"To find frequency density, divide class width by frequency density.","correct":false,"explanation":"This uses frequency density within its own calculation and divides in the wrong direction. The required calculation is frequency divided by class width."}]},{"id":"0d8f6712-cfe5-4337-bca1-92a5cc656034","cardType":"basic","points":2,"front":"A class $0 \\le t < 4$ has frequency 10. Find the frequency density.","back":"2.5.","mcqOptions":[{"text":"2.5.","correct":true,"explanation":"Frequency density equals frequency divided by class width. The class width is $4 - 0 = 4$, so the density is $10 \\div 4 = 2.5$."},{"text":"40.","correct":false,"explanation":"Multiplying frequency by class width reverses the required operation. Frequency density is frequency divided by width, so $10 \\div 4 = 2.5$."},{"text":"6.","correct":false,"explanation":"Subtracting the class width from the frequency does not calculate density. Divide the frequency by the width: $10 \\div 4 = 2.5$."},{"text":"0.4.","correct":false,"explanation":"Dividing class width by frequency gives the reciprocal of the required density. The correct calculation is $10 \\div 4$, giving a frequency density of $2.5$."}]},{"id":"8cd6de90-cd2f-4333-ba31-45819d9ef7a6","cardType":"basic","points":2,"front":"A class $4 \\le t < 9$ has frequency density 1.4. Find the frequency.","back":"7.","mcqOptions":[{"text":"7.","correct":true,"explanation":"Frequency equals frequency density multiplied by class width. The width is $9 - 4 = 5$, so the frequency is $1.4 \\times 5 = 7$."},{"text":"0.28.","correct":false,"explanation":"Dividing the density by the class width uses the operations in the wrong direction. Frequency is density multiplied by width, so $1.4 \\times 5 = 7$."},{"text":"5.6.","correct":false,"explanation":"Using the lower class boundary as the width gives $1.4 \\times 4$, but the width is $9 - 4 = 5$. Therefore, the frequency is $7$."},{"text":"12.6.","correct":false,"explanation":"Using the upper class boundary as the width gives $1.4 \\times 9$. The actual class width is $9 - 4 = 5$, so the frequency is $7$."}]},{"id":"a2f8d2f7-ca5f-4224-8356-83ae369bd41b","cardType":"basic","points":1,"front":"When drawing a histogram, should the bars touch?","back":"Yes - bars touch for continuous grouped data.","mcqOptions":[{"text":"Yes - bars touch for continuous grouped data.","correct":true,"explanation":"A histogram represents continuous grouped data, so adjacent class intervals meet without gaps. Therefore, the corresponding bars touch unless an interval has zero frequency."},{"text":"No - bars have gaps for continuous grouped data.","correct":false,"explanation":"Gaps are normally used in bar charts for separate categories or discrete values. Histograms represent continuous grouped intervals, so adjacent bars should touch."},{"text":"No - bars touch only for discrete ungrouped data.","correct":false,"explanation":"This reverses the usual distinction between chart types. Histogram bars touch because the grouped intervals form a continuous scale, while discrete bar charts normally have gaps."},{"text":"No - bars must be separated when class widths differ.","correct":false,"explanation":"Unequal class widths change the widths and frequency densities of the bars, not whether they touch. Adjacent continuous class intervals are still drawn without gaps."}]},{"id":"fc395192-8582-4b67-b12b-574cb38f2878","cardType":"basic","points":1,"front":"Why must a class from 15 to 25 be drawn wider than a class from 4 to 9?","back":"It has a larger class width.","mcqOptions":[{"text":"It has a larger class width.","correct":true,"explanation":"Histogram bar widths represent class widths on the horizontal axis. The interval from 15 to 25 has width 10, while the interval from 4 to 9 has width 5."},{"text":"It has a larger frequency.","correct":false,"explanation":"Frequency determines a bar’s area, not its horizontal width. The interval from 15 to 25 is wider because its class width is 10 rather than 5."},{"text":"It has a larger frequency density.","correct":false,"explanation":"Frequency density determines a histogram bar’s height, not its width. The bar is wider because $25 - 15 = 10$, whereas $9 - 4 = 5$."},{"text":"It has a larger upper boundary.","correct":false,"explanation":"A larger upper boundary alone does not determine bar width; the lower boundary must also be considered. The class widths are 10 and 5, respectively."}]},{"id":"bb4766a5-fc60-42a7-a2ac-6bc444d252f9","cardType":"cloze","points":null,"front":"Partial-bar estimates assume the data is [ ] within the class interval.","back":"Partial-bar estimates assume the data is evenly spread within the class interval.","mcqOptions":[{"text":"evenly spread","correct":true,"explanation":"Estimating part of a histogram bar assumes observations are distributed uniformly throughout that class. The required proportion of the class width can then be applied to its frequency."},{"text":"concentrated centrally","correct":false,"explanation":"A partial-bar estimate cannot assume values are gathered near the class midpoint. It assumes the data is evenly spread, allowing frequency to be estimated in proportion to width."},{"text":"clustered at the lower end","correct":false,"explanation":"Clustering near the lower boundary would make proportional estimates unreliable. The method assumes the data is evenly spread throughout the whole class interval."},{"text":"clustered at the upper end","correct":false,"explanation":"Clustering near the upper boundary is not the assumption used for partial bars. The estimate treats the data as evenly spread across the class interval."}]},{"id":"40bbfbe4-d3e0-43dc-afc6-31c7c6947253","cardType":"basic","points":2,"front":"A bar covers $10 \\le t < 20$ with frequency density 2.5. Estimate the frequency for $12 \\le t < 18$.","back":"15.","mcqOptions":[{"text":"15.","correct":true,"explanation":"The requested interval has width $18 - 12 = 6$. Multiplying this by the frequency density gives the estimated frequency: $6 \\times 2.5 = 15$."},{"text":"25.","correct":false,"explanation":"This uses the full class width of 10 and therefore finds the frequency for the entire bar. The requested width is 6, so the estimate is $6 \\times 2.5 = 15$."},{"text":"12.","correct":false,"explanation":"This effectively uses a density of 2 instead of the given 2.5. The interval width is 6, so the correct estimate is $6 \\times 2.5 = 15$."},{"text":"6.","correct":false,"explanation":"This gives only the width of the requested interval and ignores the frequency density. Multiply the width 6 by 2.5 to obtain an estimated frequency of 15."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Histograms","title":"Histograms Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"e0340d54-a293-44a9-a6ee-c03d1fd68878","walled":"$L38"}],"$L39","$L3a"]