Using the table, calculate the median, mode, and mean shoe size of the girls.

Mathematics
Using the table, calculate the median, mode, and mean shoe size of the girls.

ScanSolve AI Answer

Section A

Frequency distribution table:

Marks(xi)fi131144152163171N=∑fi=11\begin{array}{|c|c|} \hline Marks (x_i) & f_i \\ \hline 13 & 1 \\ 14 & 4 \\ 15 & 2 \\ 16 & 3 \\ 17 & 1 \\ \hline N = \sum f_i = 11 & \\ \hline \end{array}

Step 1: Mean

Formula: xˉ=∑fixiN\bar{x} = \frac{\sum f_i x_i}{N}

Compute ∑fixi\sum f_i x_i:

1×13=134×14=562×15=303×16=481×17=17∑fixi=164\begin{align*} 1 \times 13 &= 13 \\ 4 \times 14 &= 56 \\ 2 \times 15 &= 30 \\ 3 \times 16 &= 48 \\ 1 \times 17 &= 17 \\ \hline \sum f_i x_i &= 164 \end{align*}

Substitute: xˉ=16411\bar{x} = \frac{164}{11}

Step 2: Mode

Mode is the value of xix_i with the highest frequency.
Highest fi=4f_i = 4 at xi=14x_i = 14.
Mode = 14

Step 3: Median

Cumulative frequency (cf) table:

xificf1311144515271631017111\begin{array}{|c|c|c|} \hline x_i & f_i & cf \\ \hline 13 & 1 & 1 \\ 14 & 4 & 5 \\ 15 & 2 & 7 \\ 16 & 3 & 10 \\ 17 & 1 & 11 \\ \hline \end{array}

N=11N=11 (odd), median is the N+12=6\frac{N+1}{2} = 6th observation.
5<6≤75 < 6 \leq 7, so median =15= 15.
Median = 15

Section A: Mean = \dfrac{164}{11}, Median = 15, Mode = 14

Section B

Frequency distribution table:

Marks(xi)fi106113122131N=∑fi=12\begin{array}{|c|c|} \hline Marks (x_i) & f_i \\ \hline 10 & 6 \\ 11 & 3 \\ 12 & 2 \\ 13 & 1 \\ \hline N = \sum f_i = 12 & \\ \hline \end{array}

Step 1: Mean

Formula: xˉ=∑fixiN\bar{x} = \frac{\sum f_i x_i}{N}

Compute ∑fixi\sum f_i x_i:

6×10=603×11=332×12=241×13=13∑fixi=130\begin{align*} 6 \times 10 &= 60 \\ 3 \times 11 &= 33 \\ 2 \times 12 &= 24 \\ 1 \times 13 &= 13 \\ \hline \sum f_i x_i &= 130 \end{align*}

Substitute: xˉ=13012=656\bar{x} = \frac{130}{12} = \frac{65}{6}

Step 2: Mode

Mode is the value of xix_i with the highest frequency.
Highest fi=6f_i = 6 at xi=10x_i = 10.
Mode = 10

Step 3: Median

Cumulative frequency (cf) table:

xificf106611391221113112\begin{array}{|c|c|c|} \hline x_i & f_i & cf \\ \hline 10 & 6 & 6 \\ 11 & 3 & 9 \\ 12 & 2 & 11 \\ 13 & 1 & 12 \\ \hline \end{array}

N=12N=12 (even), median is average of N2=6\frac{N}{2}=6th and 77th observations.
66th: cf 6≥66 \geq 6, so 1010.
77th: cf 6<7≤96 < 7 \leq 9, so 1111.
Median =10+112=10.5=212= \frac{10 + 11}{2} = 10.5 = \frac{21}{2}.
Median = \dfrac{21}{2}

Section B: Mean = \dfrac{65}{6}, Median = \dfrac{21}{2}, Mode = 10

Section A: mean = 164/11, median = 15, mode = 14

Section B: mean = 65/6, median = 21/2, mode = 10

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