Histograms Flashcards

All 26 cards in this deck

What is frequency density?

Frequency ÷\div class width.

How do you work out the class width of a group?

e.g. 20<w≤5020 < w \le 50

Upper bound −- lower bound.

e.g. 50−20=3050 - 20 = 30

How do you get the frequency of a class from its frequency density?

e.g. frequency density 2.52.5, class 70<w≤9070 < w \le 90

Frequency == frequency density ×\times class width.

e.g. 2.5×20=502.5 \times 20 = 50

What does the area of a bar in a histogram represent?

The frequency of that class.

True or false? In a histogram, the height of each bar shows the frequency of that class.

False. The height is the frequency density; the area of the bar shows the frequency.

True or false? A frequency density must be a whole number.

False. It is frequency ÷\div class width, so it can be a decimal, e.g. 50÷20=2.550 \div 20 = 2.5.

What are the steps to draw a histogram from a grouped frequency table?

e.g. 0<t≤100 < t \le 10 has frequency 2020

  1. Find each class width: 10−0=1010 - 0 = 10
  2. Frequency ÷\div class width == frequency density: 20÷10=220 \div 10 = 2
  3. Draw a bar from 00 to 1010 of height 22, axis labelled 'frequency density', with no gaps between bars.

What must the vertical axis of a histogram be labelled?

Frequency density.

In a histogram, what does the width of each bar represent?

The class interval, drawn from its lower bound to its upper bound.

How do you find the height of a missing bar when completing a histogram?

e.g. the class 40<t≤6040 < t \le 60 has frequency 2424

Divide the frequency by the class width to get the frequency density.

e.g. 24÷20=1.224 \div 20 = 1.2

True or false? A histogram is drawn with gaps between the bars.

False. The data is continuous, so the bars touch each other.

True or false? All the bars of a histogram must be the same width.

False. Unequal class intervals give bars of different widths; the frequency density scale keeps the areas correct.

How do you read the frequency of a class from a histogram?

e.g. a bar of height 1.51.5 over 10<t≤2010 < t \le 20

Frequency density ×\times class width (the area of the bar).

e.g. 1.5×10=151.5 \times 10 = 15

How do you estimate the frequency in part of a class interval?

e.g. 150<h≤180150 < h \le 180 has frequency 1818; estimate for 150<h≤160150 < h \le 160

Assume the data is spread evenly and take that fraction of the class frequency.

e.g. 1030×18=6\frac{10}{30} \times 18 = 6

In a histogram, what does one bar having a greater frequency density than another tell you?

That interval contains more values per unit of the quantity, so the data is more concentrated there.

What are the steps to estimate the median from a histogram?

e.g. a histogram showing 4343 runners' times

  1. Find the total frequency: 4343
  2. Find the median position: 43÷2=21.543 \div 2 = 21.5
  3. Add bar frequencies until you reach it, then use proportion across that bar: e.g. 16+9.516×4=18.37516 + \frac{9.5}{16} \times 4 = 18.375

For nn values shown in a histogram, which position do you use to estimate the median?

The n2\frac{n}{2} th value (using n+12\frac{n+1}{2} is also accepted).

For nn values shown in a histogram, which positions do you use to estimate the lower and upper quartiles?

The n4\frac{n}{4} th and 3n4\frac{3n}{4} th values.

e.g. for n=23n = 23: 5.755.75 and 17.2517.25

What is the formula for the interquartile range?

IQR=upper quartile−lower quartile\text{IQR} = \text{upper quartile} - \text{lower quartile}

How do you find the median value once you know which bar it lies in?

e.g. 1010 more values needed into a bar of frequency 1616 covering 16<t≤2016 < t \le 20

Go that fraction of the way across the interval and add it to the lower bound.

e.g. 16+1016×4=18.516 + \frac{10}{16} \times 4 = 18.5

True or false? The median of data in a histogram is the midpoint of the interval with the tallest bar.

False. You find the interval containing the n2\frac{n}{2} th value using the bar areas, then use proportion within it.

True or false? If all the class intervals in a histogram have the same width, the heights of the bars are proportional to the frequencies.

True. Each frequency is divided by the same class width, so the frequency densities are in the same proportion as the frequencies.

What are the steps to estimate the fraction of the data in a given range from a histogram?

e.g. 150150 students in total; for 150<h≤170150 < h \le 170 the whole bars have frequencies 3030, 5151 and 3636, and one third of a bar of frequency 1818 lies in the range

  1. Add the frequencies of the whole bars in the range: 30+51+36=11730 + 51 + 36 = 117
  2. Add the proportional part of any part-bar: 13×18=6\frac{1}{3} \times 18 = 6, giving 123123
  3. Write this over the total frequency: 123150\frac{123}{150}

True or false? To estimate the proportion of the data in a given range of a histogram, you divide the number of bars in that range by the total number of bars.

False. You divide the frequency (area) in the range by the total frequency (total area).

What are the steps to estimate a percentage of the data in part of a histogram?

e.g. over 5050 years the bars have frequency density 1.41.4 for a width of 1010 and 0.70.7 for a width of 3030; 20%20\% of these members are female

  1. Find each frequency as area: 1.4×10=141.4 \times 10 = 14 and 0.7×30=210.7 \times 30 = 21
  2. Add them: 14+21=3514 + 21 = 35 members over 5050
  3. Take the percentage: 20%20\% of 35=735 = 7 female members

What are the steps to estimate the value needed to be in the top 20%20\% of a histogram?

e.g. total frequency 6060; the highest bar covers scores 3030 to 4040 and has frequency 1212

  1. Find the number needed: 20%20\% of 60=1260 = 12 people
  2. Work down from the highest values, adding frequencies until you reach 1212: the 3030–4040 bar alone gives 1212
  3. Read off the value at that point: a score of 3030