Histograms: past paper questions & worked solutions

Use \(\text{frequency density}=\text{frequency}/\text{class width}\): \[14/20=0.7,\quad12/5=2.4,\quad32/10=3.2,\quad30/15=2.\]

Histogram areas, rather than heights alone, represent frequencies.

Draw four adjoining rectangles on the supplied time and frequency-density axes: \([0,20]\) at height \(0.7\), \([20,25]\) at height \(2.4\), \([25,35]\) at height \(3.2\), and \([35,50]\) at height \(2\).

Each rectangle starts at density 0 and has the stated class boundaries as its vertical edges; leave no gaps between classes.

So the other half is also \(42\) parcels.

Frequency density \(=\dfrac{\text{frequency}}{\text{class width}}\).

For a histogram,

For \(10<t\leq15\), the frequency is 20 and the class width is 5, so

Using this scale from the histogram:

For trains at least 25 minutes late:

Total estimated frequency:

More than \(40\) minutes:

So the proportion is

Frequency is class width multiplied by frequency density.

Total:

More than \(40\) minutes is the part from \(40\) to \(50\), plus \(50\) to \(60\).

So the proportion is

The interval \(10\) to \(25\) has width \(15\) and frequency \(60\), so its frequency density is

Using that scale from the histogram:

For a histogram,

The bar heights are:

For a histogram,

For \(15\leq d<20\), the frequency is 14 and the class width is 5, so

Using this scale from the histogram:

So the known frequencies are

There are 100 adults in total, so the missing frequency from 5 km to 15 km is

The missing class width is 10, so its frequency density is

The \(0\) to \(20\) gram class has frequency \(24\), so its density is

From the histogram scale, the density for \(30<w\le 50\) is about \(2.2\), and the density for \(50<w\le 65\) is about \(1.8\).

Medium oranges weigh between \(35\) g and \(55\) g:

Use relative densities from the histogram:

The number of trees with \(300<h\le 400\) is

The number of trees with \(400<h\le 500\) is

The difference is \(8\):

For height greater than \(500\):

For a histogram,

For \(m\leq20\), the frequency is 10 and the class width is 20, so

Using this scale from the histogram:

For \(50<m\leq80\), use part of each relevant bar:

For a histogram,

For \(8\leq w<8.5\), the frequency is 16 and the class width is \(0.5\), so

Using this scale from the histogram, the frequency densities are:

For \(0<m\le 20\), the frequency is \(10\), so

Using the same scale from the histogram:

For a histogram,

For books weighing less than 100 g, the frequency is 75 and the class width is 100, so

Using this scale from the histogram, the frequency densities are:

The total number of books is

For weights between 400 g and 600 g:

So the probability is

Estimate the frequencies from each bar:

Total frequency:

Between \(20\) and \(60\) minutes:

The total frequency is \(140\).

So the two missing frequencies have total

From the histogram, the missing frequencies are

The scale of the frequency density axis is not needed because the percentage is found from areas.

Using the bar heights from the histogram, the relative areas are:

For ages more than 55, use the part of the last bar from 55 to 75:

So the estimated percentage is

For a histogram,

For \(75<h\leq85\), the frequency is 12 and the class width is 10, so

Using this scale from the histogram, the frequency densities are:

The total number of tomato plants is

For \(h>82.5\):

So the probability is

Use the relative histogram heights.

For \(155<h\le 170\), the relative area is

This represents \(160\) students, so each relative area unit represents

Total relative area:

Estimated total:

The first class has width 15 g and frequency 9, so its density is \(9/15=0.6\). Its height is 6 small grid squares, so each vertical square represents density \(0.1\).

Frequency is class width multiplied by frequency density.

The classes below 175 g have densities \(0.6,2.9,4.2,3.4\) and widths \(15,10,5,5\), respectively.

Read the bar heights using the calibrated vertical scale.

Add the frequencies, not the heights.

Above 180 g, the frequency from 180 to 185 is \(5(2.4)=12\), from 185 to 190 is \(5(5)=25\), and from 190 to 200 is \(10(1.5)=15\).

Use only half of the 175-185 bar and all of the two heavier classes.

Restrict the denominator to the large apples.

Given \(L\le 25\), the total frequency is

For \(L\le 5\), estimate half of the first class:

The \(0\le d<10\) class has frequency \(40\), so its density is

From the histogram, the missing frequencies are

Together with the table:

For \(d>25\):

The \(0<t\le 2\) class has frequency \(28\). From the histogram, the relative heights are

So the frequencies are

Use midpoints \(1,3,4.5,5.5,7.5\).

Total frequency:

Between 17 and 35 minutes:

Given the customer spent more than 30 minutes, they are in the \(30\) to \(45\) minute class.

More than 36 minutes is \(9\) minutes out of the \(15\)-minute class:

For plants taller than \(40\text{ cm}\):

From \(40<h\le 50\):

From \(50<h\le 70\):

From \(70<h\le 80\):

For \(0.5<w\le 2\), the frequency is \(63\), so the density is

Using the histogram heights, the class frequencies are

So the total number of parcels is

For weight greater than \(2.25\text{ kg}\):

The probability that both randomly chosen records are greater than \(2.25\text{ kg}\) is