9. Statistics

For shirt colours blue, red, blue, green, red, blue, the frequencies are blue $3$, red $2$, green $1$. Their sum is $6$, matching the six observations; tally marks are the counting record, not the final numerical frequencies.

For Group A, half the values are above the median $162$. Therefore an estimate of the number above $162$ is $0.5\times136=68$. This is an estimate because a box plot summarises positions rather than listing every value.

Set P is more consistent because its interquartile range is smaller: $18$ compared with $35$. The IQR compares the spread of the middle half; the range compares the full span and is more affected by extreme values.

For $24,25,26,27,98$, the mean is $40$ but the median is $26$. The single extreme value makes the mean look much larger than most observations, so the median better represents the centre of this set.

A lower median but larger IQR means one group is typically lower yet more variable; neither statement cancels the other. Conclusions should identify the exact statistic and avoid turning a summary comparison into a claim about every individual.

For ordered data $2,4,5,6,7,8,9,11$, the median is $(6+7)\div2=6.5$. The lower half is $2,4,5,6$, so $Q_1=(4+5)\div2=4.5$; the upper half gives $Q_3=(8+9)\div2=8.5$. Thus IQR $=8.5-4.5=4$, while range $=11-2=9$.

The estimate is $280\div10=28$. Use the frequency, not the class width, as the multiplier. The same midpoint method works for grouped discrete or grouped continuous intervals when their frequencies are known.

The greatest frequency is $18$, so the modal class is $1.5<h\leq1.6$ m. Give the complete interval with its inequality signs and unit, not the frequency $18$.

For frequencies $9,14,7$, the total is $30$. The sector angles are $9\div30\times360^\circ=108^\circ$, $14\div30\times360^\circ=168^\circ$ and $7\div30\times360^\circ=84^\circ$. They sum to $360^\circ$, checking the chart.

For $13,15,21,21,28,31$, use stems $1,2,3$ and ordered leaves: $1\mid3\ 5$, $2\mid1\ 1\ 8$, $3\mid1$. The key $1\mid3=13$ fixes the place value. Every original value must appear exactly once.

Read the whole cloud of points before describing it. State the variables and direction, for example: 'as time in the shop increases, the number of items bought tends to increase.' A point far from the overall pattern is an unusual point; identify it from its coordinates.

If higher temperature is generally paired with more ice creams sold, the correlation is positive. If greater car power is generally paired with a shorter acceleration time, it is negative. Judge the trend of the whole cloud, not the slope between two selected points.

To estimate $y$ from a given $x$, start at the $x$-axis value, move vertically to the line, then move horizontally to the $y$-axis and read the scale. Reverse these moves to estimate $x$ from $y$. An estimate is approximate, so report a precision justified by the scale.

If the best-fit line for two test scores crosses near $(40,48)$, a test 1 score of $40$ gives an estimated test 2 score of about $48$. The estimate comes from the line, not from choosing the nearest plotted student.

The median divides the ordered observations into two halves. The IQR measures the spread of the middle $50\%$, so a smaller IQR means the central half is more tightly clustered. Keep graph readings consistent with the scale and report them as estimates.

If one class has width $10$ and density $4$, its area and frequency are $40$. A class of width $25$ and density $2$ has frequency $50$: it is shorter but contains more observations.

To recover a frequency from the diagram, read the density and multiply by the class width. For only part of a class, the corresponding rectangle area gives an estimate based on the distribution within that class.