The marks of a group of 19 students in a test were recorded, as the marks are shown in the stem and leaf diagram. The mode is 42 marks.

  1. ((i))

    Find the value of a and b.

    1. (a)

      Answer a = ...

    2. (b)

      b = ...

  2. ((ii))

    Find the median mark.

  3. ((iii))

    Calculate the mean and standard deviation of the marks.

    1. (a)

      Answer mean = ...

    2. (b)

      Standard Deviation = ...

  4. ((iv))

    Marks for another group of 20 students was also recorded. The results are summarized in the table below.

    Make two comparisons between the marks obtain by the group of 19 students and by the group of 20 students.

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Worked solution

4((i))(a)

1. Use the given mode to restrict the unknown leaf $a$.

The mark 4242 appears 33 times.

Since the mode is 4242, the unknown mark 3a3a must not create another mark appearing 33 times. In the stem 3 row, the leaves are
a, 0, 1, 2, 6.a,\ 0,\ 1,\ 2,\ 6.
For the leaves to remain in order, aa must be before 00, so
a=0.a=0.

Answer

a=0a=0

4((i))(b)

1. Determine the missing leaf from the ordered leaves in the stem 5 row.

In the stem 55 row, the leaves are
0, 2, 4, 4, b, 5, 8, 90,\ 2,\ 4,\ 4,\ b,\ 5,\ 8,\ 9.

Since leaves are written in ascending order,
4≤b≤5.4 \leq b \leq 5.
As bb is a single-digit leaf and is placed before 55 after two 44s, the missing leaf is
b=4.b=4.

Answer

b=4b=4

4((ii))

1. Locate the median position for 19 marks.

There are 1919 students.

Median position =19+12=10=\dfrac{19+1}{2}=10.

So the median is the 1010th mark in ascending order.

2. Read the 10th mark from the ordered stem-and-leaf diagram using the earlier values of $a$ and $b$.

Using a=0a=0 and b=4b=4, the marks in order are

30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,5930,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59.

The 1010th mark is 4242.

Answer

4242 marks

4((iii))(a)

1. Use all 19 marks to calculate the mean.

Using a=0a=0 and b=4b=4, the marks are
30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,5930,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59.

Mean=30+30+31+32+36+41+42+42+42+46+47+50+52+54+54+54+55+58+5919\text{Mean}=\frac{30+30+31+32+36+41+42+42+42+46+47+50+52+54+54+54+55+58+59}{19}

=85519=45=\frac{855}{19}=45

Answer

4545 marks

4((iii))(b)

1. Use the completed list of marks and calculate the standard deviation from the mean.

The 19 marks are
30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59.30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59.
Using the mean from part (iii)(a), xˉ=45\bar{x}=45,
σ=∑(x−45)219.\sigma=\sqrt{\frac{\sum (x-45)^2}{19}}.
Now
∑(x−45)2=1216.\sum (x-45)^2=1216.
So
σ=121619=64=8.\sigma=\sqrt{\frac{1216}{19}}=\sqrt{64}=8.

Answer

8.008.00 marks

4((iv))

1. Compare the average marks using the means.

For the group of 19 students, mean =45=45 marks.

For the group of 20 students, mean =48.2=48.2 marks.

Since 48.2>4548.2>45, the group of 20 students obtained higher marks on average.

2. Compare the spread or consistency of the marks using the standard deviations.

For the group of 19 students, standard deviation =8.00=8.00 marks.

For the group of 20 students, standard deviation =7.22=7.22 marks.

Since 7.22<8.007.22<8.00, the group of 20 students had less spread in its marks, so its marks were more consistent.

Answer

The group of 20 students did better on average since its mean mark is 48.248.2, which is higher than 4545 for the group of 19 students.

The group of 19 students had more consistent marks since its standard deviation is 8.008.00, which is lower than 7.227.22?

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