Given a table of data, calculate the mean, mode and median, and complete a frequency table.
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\n| Number of siblings | \nFrequency | \n
|---|---|
| $0$ | \n[[0]] | \n
| $1$ | \n[[1]] | \n
| $2$ | \n[[2]] | \n
| $3$ | \n[[3]] | \n
| $4$ | \n[[4]] | \n
| $5$ | \n[[5]] | \n
| $6$ | \n[[6]] | \n
| Total | \n$30$ | \n
Find the mean, mode and median for this data.
\nMean = [[0]]
\nMode = [[1]]
\nMedian = [[2]]
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\nEach row of the frequency column gives the number of students with the corresponding number of siblings.
\n| Number of siblings | \nFrequency | \n
|---|---|
| $0$ | \n$\\var{freq[0]}$ | \n
| $1$ | \n$\\var{freq[1]}$ | \n
| $2$ | \n$\\var{freq[2]}$ | \n
| $3$ | \n$\\var{freq[3]}$ | \n
| $4$ | \n$\\var{freq[4]}$ | \n
| $5$ | \n$\\var{freq[5]}$ | \n
| $6$ | \n$\\var{freq[6]}$ | \n
| Total | \n$30$ | \n
Always remember to check whether your frequency column adds up to the total (here, it is $30$) to make sure you have not left out any responses.
\nThe mean number of siblings is the total number of siblings, $\\sum x$, divided by the number of students in the sample, $n$.
\n\\begin{align}
\\sum x &= 0 \\times \\var{freq[0]} + 1 \\times \\var{freq[1]} + 2 \\times \\var{freq[2]} + 3 \\times \\var{freq[3]} + 4 \\times \\var{freq[4]} + 5 \\times \\var{freq[5]} + 6 \\times \\var{freq[6]}
\\\\
&= 0 + \\var{1*freq[1]} + \\var{2*freq[2]} + \\var{3*freq[3]} + \\var{4*freq[4]} + \\var{5*freq[5]} + \\var{6*freq[6]} \\\\&= \\var{sum(a)} \\text{.}
\\end{align}
The total number of students $n$ is $30$.
\nTherefore the mean is
\n\\begin{align}
\\bar{x} &= \\frac{\\sum x}{n} \\\\
&= \\frac{\\var{sum(a)}}{30} \\\\
&= \\var{mean} \\text{.}
\\end{align}
Rounding the answer to 2 decimal places, we get $\\var{precround(mean, 2)}$.
\nThe mode is the value with the highest frequency. Here, the mode is $\\var{mode}$ siblings, with frequency $\\var{freq[mode]}$.
\nThe median is the \"middle\" value in the sample, when arranged in numerical order.
\nSince $n = 30$, we have two middle values in this data (15th and 16th place). We can count from the top of the table until we locate rows where these middle values lie, as the numbers in the table are already sorted by order.
\nHere, both $15$th and $16$th value lie in the row $\\var{as[14]}$.Here, the $15$th value lies in the row $\\var{as[14]}$ while the $16$th value lies in the row $\\var{as[15]}$.
\nAs $15$th value $= 16$th value $= \\var{as[14]}$, the median is $\\var{as[14]}$.As $15$th value $= \\var{as[14]}$ and $16$th value $= \\var{as[15]}$, we need to find their mean.
\n\\[ \\displaystyle \\begin{align} \\frac{\\var{as[14]} + \\var{as[15]}}{2} &= \\frac{\\var{as[14] + as[15]}}{2} \\\\&= \\var{median} \\text{.} \\end{align}\\]
\nThis is the median for this data.
\n", "statement": "30 random students were asked about the number of siblings they have. These are their responses:
\n| $\\var{a[0]}$ | \n$\\var{a[1]}$ | \n$\\var{a[2]}$ | \n$\\var{a[3]}$ | \n$\\var{a[4]}$ | \n$\\var{a[5]}$ | \n$\\var{a[6]}$ | \n$\\var{a[7]}$ | \n$\\var{a[8]}$ | \n$\\var{a[9]}$ | \n
| $\\var{a[10]}$ | \n$\\var{a[11]}$ | \n$\\var{a[12]}$ | \n$\\var{a[13]}$ | \n$\\var{a[14]}$ | \n$\\var{a[15]}$ | \n$\\var{a[16]}$ | \n$\\var{a[17]}$ | \n$\\var{a[18]}$ | \n$\\var{a[19]}$ | \n
| $\\var{a[20]}$ | \n$\\var{a[21]}$ | \n$\\var{a[22]}$ | \n$\\var{a[23]}$ | \n$\\var{a[24]}$ | \n$\\var{a[25]}$ | \n$\\var{a[26]}$ | \n$\\var{a[27]}$ | \n$\\var{a[28]}$ | \n$\\var{a[29]}$ | \n