Interpreting Cumulative Frequency Diagrams Flashcards
All 9 cards in this deck
How do you estimate the median from a cumulative frequency graph?
e.g. a graph for studentsRead across from cumulative frequency to the curve, then down to the horizontal axis.
At which cumulative frequencies do you read off the lower and upper quartiles?
e.g. a graph for studentsLower quartile at and upper quartile at ; read across to the curve then down.
How do you estimate the number of values greater than a given value from a cumulative frequency graph?
e.g. of students, how many are taller than cmRead the cumulative frequency at that value and subtract it from the total: if the reading at is , the answer is .
How do you estimate the number of values between two given values from a cumulative frequency graph?
e.g. between g and gRead the cumulative frequency at each value and subtract: if the readings are at g and at g, the answer is .
How do you estimate the 90th percentile from a cumulative frequency graph?
e.g. a graph for itemsRead across at of the total, , then down to the horizontal axis.
What must your answer include when asked whether a claim about a cumulative frequency graph or box plot is valid?
A decision (yes or no) and a reason quoting values read from the graph.
e.g. 'No, the median is g, not g.'True or false? The median and quartiles found from a cumulative frequency graph are only estimates.
True. The original data is grouped, so exact values are unknown and readings from the curve are estimates.
How do you use a cumulative frequency graph to estimate the value that a given percentage of the data lies below?
e.g. the weight below which the lightest 25% of 60 potatoes lie
- Find that percentage of the total: 25% of 60 = 15
- Read across from cumulative frequency 15 to the curve, then down to the weight axis
- That weight is the estimate
How do you estimate what percentage of the data is below a given value on a cumulative frequency graph?
e.g. the percentage of 80 plants shorter than 90 cm
Read the cumulative frequency at that value, then divide by the total and multiply by 100.
e.g. reading 74 gives