Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — mean and standard deviation of a short list
Find the mean and the standard deviation of 2,4,4,5,7,8.
1∑x=2+4+4+5+7+8=30, and n=6Total first, then count.
2xˉ=630=5The mean.
3∑x2=4+16+16+25+49+64=174Square each value, then add. Not the square of the total.
4Variance =6174−52=29−25=4n∑x2−xˉ2.
5Standard deviation =4=2Take the root. Stopping at 4 answers the wrong question.
Example 2 — working from a frequency table
For the table below, find the mean and the standard deviation.
x
2
4
6
8
f
1
3
4
2
1∑f=1+3+4+2=10The number of items, not the number of columns.
2∑fx=2+12+24+16=54Multiply each x by its own frequency, then total.
3xˉ=1054=5.4Divide by ∑f.
4∑fx2=1(4)+3(16)+4(36)+2(64)=4+48+144+128=324Square the x first, then multiply by f.
5Variance =10324−5.42=32.4−29.16=3.24
6Standard deviation =3.24=1.8A tidy answer, which is usually a sign the arithmetic held together.
Example 3 — quartiles from a cumulative frequency curve
A cumulative frequency curve is drawn for the marks of 200 students. Reading from the curve gives Q1=42, Q2=55 and Q3=64. Find the interquartile range and the semi-interquartile range, and state how many students scored above 64.
1The three readings are taken at cumulative frequencies of 50, 100 and 150.4n, 2n and 43n with n=200.
2Interquartile range =Q3−Q1=64−42=22The width of the middle half of the data.
3Semi-interquartile range =222=11Half the interquartile range — read the question for which one is wanted.
4Q3 is at a cumulative frequency of 150, so 150 students scored 64 or less.That is what the upper quartile means.
5200−150=50 students scored above 64.A quarter of them, which is exactly what the upper quartile marks off.