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The scores total 150, so dividing by 10 gives the correct mean of 15 marks."},{"text":"14.5 marks","correct":false,"explanation":"This near-miss may come from miscounting the scores or the total. There are ten scores adding to 150, so the mean is 15 marks."},{"text":"16 marks","correct":false,"explanation":"This overestimates the average, perhaps by treating the largest scores as more important. The total is 150 across ten scores, giving a mean of 15 marks."}]},{"id":"1da00017-e893-4eda-955b-431bada5d98e","cardType":"basic","points":1,"front":"How do you find the median when there is an even number of scores?","back":"For an even number of scores, average the two middle values.","mcqOptions":[{"text":"For an even number of scores, average the two middle values.","correct":true,"explanation":"With an even number of ordered scores, no single score occupies the middle position. Averaging the two central values gives the median."},{"text":"For an even number of scores, choose the single middle value.","correct":false,"explanation":"A single middle score applies when there is an odd number of scores. With an even number, the two middle values must be averaged."},{"text":"For an even number of scores, average the two end values.","correct":false,"explanation":"The end values are the minimum and maximum and do not identify the centre. The median comes from averaging the two middle values."},{"text":"For an even number of scores, choose the most frequent value.","correct":false,"explanation":"The most frequent value is the mode, not the median. For an even number of ordered scores, calculate the median by averaging the two middle values."}]},{"id":"b4c1586a-96e9-43ea-9262-20f60f459519","cardType":"basic","points":1,"front":"With seven ordered scores, which position is the median?","back":"The median is the 4th score.","mcqOptions":[{"text":"The median is the 4th score.","correct":true,"explanation":"For seven ordered scores, the middle position is found with (7 + 1) divided by 2, which equals the 4th score."},{"text":"The median is the 3rd score.","correct":false,"explanation":"The 3rd score leaves three scores on each side only if the centre score is counted incorrectly. For seven scores, the median is the 4th score."},{"text":"The median is the 5th score.","correct":false,"explanation":"The 5th score is one position to the right of the centre. Seven ordered scores have their median at the 4th score."},{"text":"The median is the 7th score.","correct":false,"explanation":"The 7th score is the highest value, not the centre. In an ordered set of seven scores, the median is the 4th score."}]},{"id":"145955ab-b5ba-4c48-a2eb-efcd9833f298","cardType":"basic","points":1,"front":"What is the mode of 11, 13, 13, 14, 15, 15, 15, 16, 18 and 20?","back":"15 marks","mcqOptions":[{"text":"15 marks","correct":true,"explanation":"The mode is the value occurring most often. Here, 15 appears three times, more frequently than any other score."},{"text":"13 marks","correct":false,"explanation":"Although 13 occurs twice, it is not the most frequent value. The score 15 appears three times and is therefore the mode."},{"text":"14 marks","correct":false,"explanation":"The value 14 appears only once, so it cannot be the mode. The most frequent score is 15, which appears three times."},{"text":"16 marks","correct":false,"explanation":"The value 16 appears once and is not the most common score. The mode is 15 because it occurs three times."}]},{"id":"37ebe6c1-86b5-40fe-a34e-4a6cd47bde81","cardType":"basic","points":1,"front":"What is reported as the mode when data is grouped into bands?","back":"The band with the highest frequency.","mcqOptions":[{"text":"The band with the highest frequency.","correct":true,"explanation":"For grouped data, individual repeated values are unavailable. The modal group is therefore the band containing the greatest number of scores."},{"text":"The band with the lowest frequency.","correct":false,"explanation":"The least common band does not represent the mode. The mode for grouped data is the band with the highest frequency."},{"text":"The widest band.","correct":false,"explanation":"Band width describes the interval size, not how common scores are. The modal band is identified by having the highest frequency."},{"text":"The middle band.","correct":false,"explanation":"The central band is not necessarily the most common. Grouped data reports the band with the highest frequency as the mode."}]},{"id":"ba0b9db4-cf39-4a24-be0c-978df30bf64b","cardType":"basic","points":1,"front":"What is the range of scores from 11 to 20?","back":"9 marks","mcqOptions":[{"text":"9 marks","correct":true,"explanation":"The range measures the distance between the highest and lowest scores: 20 minus 11 equals 9 marks."},{"text":"10 marks","correct":false,"explanation":"Counting both endpoints gives ten score values, but range is a difference rather than a count. Subtracting 11 from 20 gives 9 marks."},{"text":"31 marks","correct":false,"explanation":"Adding the endpoints confuses total with spread. Range is calculated by subtracting the lowest score from the highest: 20 - 11 = 9 marks."},{"text":"11 marks","correct":false,"explanation":"This is the lowest score, not the distance between the extremes. The range is found by 20 - 11, giving 9 marks."}]},{"id":"92d9323b-7627-4e7f-b816-49e37d1c6108","cardType":"basic","points":1,"front":"In a fraction, where are the part and the whole written?","back":"The part is written on top and the total underneath.","mcqOptions":[{"text":"The part is written on top and the total underneath.","correct":true,"explanation":"A fraction represents a selected part compared with the whole. The numerator is the part and the denominator is the total."},{"text":"The total is written on top and the part underneath.","correct":false,"explanation":"This reverses numerator and denominator. To represent a part of a whole, write the part on top and the total underneath."},{"text":"The part is written underneath and the total beside it.","correct":false,"explanation":"A fraction uses a numerator above a denominator, not a side-by-side layout. The part goes on top and the total underneath."},{"text":"The difference is written on top and the total underneath.","correct":false,"explanation":"A difference is not the selected quantity being compared with the whole. The numerator should be the part, with the total as the denominator."}]},{"id":"fe8cb58c-f6d0-4bd8-8750-2fc5460b6216","cardType":"basic","points":1,"front":"How do you calculate a percentage from a part and a total?","back":"Divide the part by the total, then multiply by 100.","mcqOptions":[{"text":"Divide the part by the total, then multiply by 100.","correct":true,"explanation":"The fraction part divided by total gives a proportion. Multiplying that proportion by 100 converts it into a percentage."},{"text":"Divide the total by the part, then multiply by 100.","correct":false,"explanation":"Reversing the division produces the reciprocal proportion and usually an inflated result. The correct calculation is part divided by total, multiplied by 100."},{"text":"Divide the part by the total, then multiply by 10.","correct":false,"explanation":"Multiplying by 10 does not convert a proportion into a percentage. The conversion requires multiplication by 100 after dividing part by total."},{"text":"Add the part to the total, then multiply by 100.","correct":false,"explanation":"Adding the quantities does not express the part as a fraction of the whole. Divide the part by the total, then multiply by 100."}]},{"id":"db91f715-ca01-4a1b-bcea-a05b41461bb7","cardType":"basic","points":1,"front":"How should the order of amounts in a ratio be written?","back":"Keep the amounts in the order given in the question.","mcqOptions":[{"text":"Keep the amounts in the order given in the question.","correct":true,"explanation":"A ratio compares quantities in a specified order, so reversing them changes its meaning. The amounts must remain in the order stated in the question."},{"text":"Keep the amounts in alphabetical order.","correct":false,"explanation":"Alphabetical ordering ignores which quantity is being compared first. Ratio terms must stay in the order given in the question."},{"text":"Put the larger amount first.","correct":false,"explanation":"A ratio is not automatically arranged from largest to smallest. Changing the order can reverse the comparison, so use the order given in the question."},{"text":"Put the smaller amount first.","correct":false,"explanation":"Choosing the smaller quantity first is just as misleading as choosing the larger one first. The question determines the required order of the ratio."}]},{"id":"c39a1c1a-d92a-44f8-b353-e8386c7ecf0e","cardType":"basic","points":2,"front":"How do you calculate percentage change?","back":"Divide the change by the starting figure, then multiply by 100.","mcqOptions":[{"text":"Divide the change by the starting figure, then multiply by 100.","correct":true,"explanation":"Percentage change compares the amount of change with the original baseline. Dividing by the starting figure and multiplying by 100 gives the percentage change."},{"text":"Divide the change by the final figure, then multiply by 100.","correct":false,"explanation":"Using the final figure changes the baseline and gives a different percentage. Percentage change uses the starting figure as the reference point."},{"text":"Divide the starting figure by the change, then multiply by 100.","correct":false,"explanation":"This reverses the required ratio and can produce an exaggerated result. The change must be divided by the starting figure before multiplying by 100."},{"text":"Divide the change by the average figure, then multiply by 100.","correct":false,"explanation":"The average is not the standard baseline for percentage change. The correct denominator is the starting figure, because change is measured from the original value."}]},{"id":"a926fa58-ec2a-4c81-a9f5-144ae9190b0b","cardType":"basic","points":2,"front":"When should the mean be used as a measure of central tendency?","back":"Use the mean when scores are numerical and no value is far from the others, because it uses all the data.","mcqOptions":[{"text":"Use the mean when scores are numerical and no value is far from the others, because it uses all the data.","correct":true,"explanation":"The mean is appropriate for numerical scores without extreme values. It incorporates every score and provides a useful measure of the centre."},{"text":"Use the median when scores are numerical and no value is far from the others, because it uses all the data.","correct":false,"explanation":"The median is based on position and does not use every score in its calculation. For numerical, evenly distributed scores, the mean uses all the data."},{"text":"Use the mode when scores are numerical and no value is far from the others, because it uses all the data.","correct":false,"explanation":"The mode identifies only the most frequent value and does not use every score. The mean is suitable when numerical scores have no extreme values."},{"text":"Use the range when scores are numerical and no value is far from the others, because it uses all the data.","correct":false,"explanation":"The range measures spread using only the highest and lowest scores, not all the data. The mean is the central-tendency measure that uses every score."}]},{"id":"309df9db-cbb7-472a-b2f6-f9ccef5541c6","cardType":"basic","points":2,"front":"Why is the median better than the mean when there is an extreme score?","back":"Use the median when an extreme score would pull the mean away from the typical value.","mcqOptions":[{"text":"Use the median when an extreme score would pull the mean away from the typical value.","correct":true,"explanation":"The median depends on the ordered position of scores, so an extreme value has little effect. It can therefore better represent the typical score than the mean."},{"text":"Use the mean when an extreme score would pull the mean away from the typical value.","correct":false,"explanation":"An extreme score directly affects the mean, which is why it becomes misleading in this situation. The median is less influenced and better represents the centre."},{"text":"Use the mode when an extreme score would pull the mean away from the typical value.","correct":false,"explanation":"The mode may be useful for the most common value, but the key protection against an extreme score is the median's resistance to outliers."},{"text":"Use the range when an extreme score would pull the mean away from the typical value.","correct":false,"explanation":"The range is a measure of spread and is itself strongly affected by extreme values. The median is the suitable central measure in this situation."}]},{"id":"1426ad1a-b48f-4aef-a5f7-380d300d129d","cardType":"basic","points":2,"front":"Why is the mode suitable for data such as favourite subject?","back":"Use the mode for categorical data because categories cannot be added together.","mcqOptions":[{"text":"Use the mode for categorical data because categories cannot be added together.","correct":true,"explanation":"Favourite subject is categorical rather than numerical, so calculating a mean or median is inappropriate. The mode identifies the category chosen most often."},{"text":"Use the mean for categorical data because categories cannot be added together.","correct":false,"explanation":"The inability to add categories makes the mean unsuitable, not suitable. The mode works because it identifies the most frequently occurring category."},{"text":"Use the median for categorical data because categories cannot be added together.","correct":false,"explanation":"Categories such as subjects cannot be meaningfully ordered to find a middle value. The mode is appropriate because it records the most common category."},{"text":"Use the range for categorical data because categories cannot be added together.","correct":false,"explanation":"The range requires numerical high and low values, which categorical labels do not provide. The mode is suitable for finding the most common category."}]},{"id":"3acda0b1-dd10-4d4d-9a7f-0cb2a8818b67","cardType":"basic","points":2,"front":"What is the main weakness of the range?","back":"It uses only the highest and lowest values, so one unusual score can stretch it.","mcqOptions":[{"text":"It uses only the highest and lowest values, so one unusual score can stretch it.","correct":true,"explanation":"The range ignores all scores between the extremes. A single unusually high or low value can therefore make the spread appear larger than is typical."},{"text":"It uses only the middle values, so one unusual score can stretch it.","correct":false,"explanation":"The range does not use middle values; it uses the two extremes. That is precisely why one unusual highest or lowest score can stretch it."},{"text":"It uses the most frequent value, so one unusual score can stretch it.","correct":false,"explanation":"The most frequent value is the mode, not the range. Range depends on the highest and lowest values, making it sensitive to an unusual score."},{"text":"It uses every score equally, so one unusual score can stretch it.","correct":false,"explanation":"The range does not use every score, so this describes the mean more closely. Its weakness is reliance on only the highest and lowest values."}]},{"id":"d10c8129-61db-41e0-898c-5e6e143aef58","cardType":"basic","points":2,"front":"Why should the range not be reported on its own?","back":"A range should be reported alongside a typical value, because spread needs context.","mcqOptions":[{"text":"A range should be reported alongside a typical value, because spread needs context.","correct":true,"explanation":"The range shows how widely scores vary but not where most scores are located. A typical value gives the spread meaningful context."},{"text":"A range should be reported alongside the highest value, because spread needs context.","correct":false,"explanation":"The highest value is already used to calculate the range and does not show the typical position of scores. A typical value provides the needed context."},{"text":"A range should be reported alongside the sample size, because spread needs context.","correct":false,"explanation":"Sample size is useful information but does not indicate where scores usually lie. The range should be paired with a typical value to interpret its spread."},{"text":"A range should be reported alongside the lowest value, because spread needs context.","correct":false,"explanation":"The lowest value is also already part of the range and does not describe the centre. A typical value shows what score is representative."}]},{"id":"2837bd7a-a088-4955-b485-e73ff7f9b7c9","cardType":"basic","points":1,"front":"What is the shape of a normal distribution?","back":"A normal distribution is bell-shaped with one peak.","mcqOptions":[{"text":"A normal distribution is bell-shaped with one peak.","correct":true,"explanation":"A normal distribution is symmetrical around its centre, with scores most common near the middle and progressively less common towards both extremes."},{"text":"A normal distribution is U-shaped with two peaks.","correct":false,"explanation":"Two peaks and high frequencies at the extremes describe a bimodal or U-shaped pattern, not a normal distribution. Normal data has one central peak."},{"text":"A normal distribution is rectangular with no peak.","correct":false,"explanation":"A rectangular shape represents roughly equal frequencies across values. A normal distribution instead has a single central peak and bell-shaped symmetry."},{"text":"A normal distribution is skewed with one tail.","correct":false,"explanation":"Skewed data has an unequal tail and lacks normal symmetry. A normal distribution is symmetrical and bell-shaped with one central peak."}]},{"id":"16b60d8d-7244-4387-b75e-996fcc4caaa1","cardType":"basic","points":1,"front":"What does symmetry mean in a normal distribution?","back":"The two halves are mirror images of each other.","mcqOptions":[{"text":"The two halves are mirror images of each other.","correct":true,"explanation":"A normal distribution is symmetrical because folding its graph vertically through the centre would make the two halves match exactly."},{"text":"The two halves are different sizes from each other.","correct":false,"explanation":"Different-sized halves describe a skewed distribution, not a symmetrical one. In a normal distribution, the two halves are mirror images of each other."},{"text":"The two halves are rotations of each other.","correct":false,"explanation":"Rotational similarity is not what symmetry means here. The two halves of a normal distribution are mirror images across a vertical line through the centre."},{"text":"The two halves are unequal shapes compared with each other.","correct":false,"explanation":"Unequal shapes would mean the distribution is asymmetrical or skewed. A normal distribution has two halves that are mirror images of each other."}]},{"id":"62296a06-445d-445e-9329-b45b2a12e14a","cardType":"basic","points":1,"front":"Where are most values found in a normal distribution?","back":"Most values cluster around the middle, with fewer at each extreme.","mcqOptions":[{"text":"Most values cluster around the middle, with fewer at each extreme.","correct":true,"explanation":"A normal distribution peaks in the middle because scores near the mean are most frequent, while increasingly extreme scores occur less often."},{"text":"Most values cluster around the extremes, with fewer in the middle.","correct":false,"explanation":"This reverses the normal distribution’s pattern. Most values cluster around the middle, and the frequency gradually decreases towards both extremes."},{"text":"Most values cluster around the lower extreme, with fewer at the upper extreme.","correct":false,"explanation":"Clustering at the lower end would produce a skewed distribution. In a normal distribution, most values cluster around the middle, with fewer at either extreme."},{"text":"Most values cluster around the upper extreme, with fewer at the lower extreme.","correct":false,"explanation":"Clustering at the upper end would indicate skew rather than normality. A normal distribution has most values near the middle and fewer at both extremes."}]},{"id":"fd3ca573-29f2-4eda-b1cd-9689381c8852","cardType":"basic","points":2,"front":"Where are the three measures of central tendency in a normal distribution?","back":"The mean, median and mode are equal at the centre under the peak.","mcqOptions":[{"text":"The mean, median and mode are equal at the centre under the peak.","correct":true,"explanation":"In a perfectly normal distribution, symmetry and a single central peak make the mean, median and mode equal at the same central point."},{"text":"The mean, median and mode are different and spread across the curve.","correct":false,"explanation":"This may occur in a skewed distribution, but not a normal one. In a normal distribution, all three measures are equal at the centre under the peak."},{"text":"The mean and median are equal, but the mode is at an extreme.","correct":false,"explanation":"Placing the mode at an extreme confuses it with an unusual score. In a normal distribution, the mean, median and mode coincide at the central peak."},{"text":"The median and mode are equal, but the mean is at an extreme.","correct":false,"explanation":"The mean can be pulled towards an extreme in a skewed distribution, but a normal distribution is symmetrical, so all three measures coincide at the centre."}]},{"id":"878f13fb-859f-4af5-a056-8e9f99541b39","cardType":"cloze","points":null,"front":"Six of ten pupils scored 15 or more, giving a fraction of $\\text{[...]}$, which simplifies to $\\frac{3}{5}$.","back":"Six of ten pupils scored 15 or more, giving a fraction of $\\boxed{\\frac{6}{10}}$, which simplifies to $\\frac{3}{5}$.","mcqOptions":[{"text":"$$\\frac{6}{10}$","correct":true,"explanation":"The fraction places the six pupils meeting the criterion over all ten pupils, giving six tenths before it is simplified to three fifths."},{"text":"$$\\frac{10}{6}$","correct":false,"explanation":"This reverses the numerator and denominator. The six qualifying pupils must be placed over the total of ten, giving six tenths."},{"text":"$$\\frac{6}{4}$","correct":false,"explanation":"Using four as the denominator confuses the fraction with the comparison between higher and lower scorers. The denominator is all ten pupils, giving six tenths."},{"text":"$$\\frac{4}{10}$","correct":false,"explanation":"This counts the four pupils scoring below 15 rather than the six scoring 15 or more. The required fraction is six tenths."}]},{"id":"6c028d9b-2652-4135-95ef-12434a5eed1d","cardType":"cloze","points":null,"front":"Six of ten pupils scored 15 or more, giving a fraction of $\\frac{6}{10}$, which simplifies to $\\text{[...]}$.","back":"Six of ten pupils scored 15 or more, giving a fraction of $\\frac{6}{10}$, which simplifies to $\\boxed{\\frac{3}{5}}$.","mcqOptions":[{"text":"$$\\frac{3}{5}$","correct":true,"explanation":"Dividing both the numerator and denominator of six tenths by their common factor of two gives three fifths, an equivalent fraction in simplest form."},{"text":"$$\\frac{6}{5}$","correct":false,"explanation":"Dividing only the denominator by two changes the fraction’s value. Both six and ten must be divided by two, producing three fifths."},{"text":"$$\\frac{2}{5}$","correct":false,"explanation":"This incorrectly changes six to two, rather than dividing both parts by the same factor. Dividing six tenths by two gives three fifths."},{"text":"$$\\frac{3}{10}$","correct":false,"explanation":"Dividing only the numerator by two halves the fraction instead of simplifying it equivalently. Both parts must be divided by two, giving three fifths."}]},{"id":"df5046ee-1f52-44f8-bea1-a9cf3c9a0317","cardType":"cloze","points":null,"front":"Six out of ten pupils scored 15 or more: $6 \\div 10 = 0.6$, so the percentage is [ ]. The ratio of higher scorers to lower scorers is $6:4$, simplifying to $3:2$.","back":"Six out of ten pupils scored 15 or more: $6 \\div 10 = 0.6$, so the percentage is 60%. The ratio of higher scorers to lower scorers is $6:4$, simplifying to $3:2$.","mcqOptions":[{"text":"60%","correct":true,"explanation":"Converting the decimal 0.6 to a percentage requires multiplying by 100, so six out of ten pupils is equal to 60%."},{"text":"6%","correct":false,"explanation":"This treats six out of ten as six out of one hundred. Multiplying the decimal 0.6 by 100 gives the correct percentage of 60%."},{"text":"0.6%","correct":false,"explanation":"This adds a percent sign without converting the decimal. A decimal is converted to a percentage by multiplying by 100, so 0.6 equals 60%."},{"text":"600%","correct":false,"explanation":"This moves the decimal point one place too far, effectively multiplying by 1,000. Multiplying 0.6 by 100 gives 60%."}]},{"id":"d1575134-cf43-4742-8507-26a5bd47e47f","cardType":"cloze","points":null,"front":"Six out of ten pupils scored 15 or more: $6 \\div 10 = 0.6$, so the percentage is 60%. The ratio of higher scorers to lower scorers is $6:4$, simplifying to $\\text{[...]}$.","back":"Six out of ten pupils scored 15 or more: $6 \\div 10 = 0.6$, so the percentage is 60%. The ratio of higher scorers to lower scorers is $6:4$, simplifying to $\\boxed{3:2}$.","mcqOptions":[{"text":"$$3:2$","correct":true,"explanation":"Both parts of the ratio six to four share a factor of two. Dividing each part by two produces the simplest ratio, three to two."},{"text":"$$2:3$","correct":false,"explanation":"This reverses the order of the two groups after simplifying. Keeping higher scorers first, dividing six to four by two gives three to two."},{"text":"$$3:4$","correct":false,"explanation":"This divides only the first part of the ratio by two. Both parts must be divided by the same number, giving three to two."},{"text":"$$6:2$","correct":false,"explanation":"This divides only the second part of the ratio by two, changing the comparison. Dividing both six and four by two gives three to two."}]},{"id":"1fa7ffe0-648e-40b3-8eda-24630cf324ce","cardType":"cloze","points":null,"front":"For revision times of 20, 25, 30, 35 and 150 minutes, the mean is [ ], while the median is 30 minutes. The median better represents the typical pupil because 150 is an extreme value.","back":"For revision times of 20, 25, 30, 35 and 150 minutes, the mean is 52 minutes, while the median is 30 minutes. The median better represents the typical pupil because 150 is an extreme value.","mcqOptions":[{"text":"52 minutes","correct":true,"explanation":"The five revision times total 260 minutes. Dividing this total by the five pupils gives a mean revision time of 52 minutes."},{"text":"65 minutes","correct":false,"explanation":"This results from dividing the 260-minute total by four instead of five. There are five values, so the mean is 260 divided by five, or 52 minutes."},{"text":"260 minutes","correct":false,"explanation":"This is the total of the five revision times, not their mean. Dividing the 260-minute total by five gives a mean of 52 minutes."},{"text":"48 minutes","correct":false,"explanation":"This can result from misadding the five revision times as 240. Their actual total is 260, which divided by five gives 52 minutes."}]},{"id":"6a2a64a0-1595-45b1-8611-428aa0dc73cd","cardType":"cloze","points":null,"front":"For revision times of 20, 25, 30, 35 and 150 minutes, the mean is 52 minutes, while the median is [ ]. The median better represents the typical pupil because 150 is an extreme value.","back":"For revision times of 20, 25, 30, 35 and 150 minutes, the mean is 52 minutes, while the median is 30 minutes. The median better represents the typical pupil because 150 is an extreme value.","mcqOptions":[{"text":"30 minutes","correct":true,"explanation":"The values are already ordered, and 30 minutes is the third of five values. It is therefore the middle value and the median."},{"text":"35 minutes","correct":false,"explanation":"This selects the value immediately above the middle. With five ordered values, the median is the third value, which is 30 minutes."},{"text":"52 minutes","correct":false,"explanation":"This confuses the median with the mean, which is increased by the extreme value of 150. The middle ordered value, and therefore the median, is 30 minutes."},{"text":"25 minutes","correct":false,"explanation":"This selects the value immediately below the middle. The third value in the ordered set is 30 minutes, so that is the median."}]}]}],"examCountdown":{"examDate":"2027-05-10","paperLabel":"Paper 1"},"heading":"11.2.2 Descriptive statistics","title":"11.2.2 Descriptive statistics Flashcards","topbar":{"backHref":"/gcse/psychology/edexcel/flashcards"},"topicSidebar":"$L90"}]}],"topicId":"8bcd3eda-676c-4990-a94f-316f7ef6de19","walled":"$L91"}],"$L92","$L93"]