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3 parts
b(i)

35.939\text{35.939}

b(ii)

46.061\text{46.061}

b(iii)

45.778\text{45.778}

Step-by-step solution

6 steps

Part a(i)

Finite set and infinite set;

Differentiate between each of the following types of sets.

  • A finite set is a set whose elements can be counted, and the counting process eventually ends. It has a definite, limited number of elements.
  • An infinite set is a set whose elements cannot be counted, as the counting process would never end. It has an unlimited number of elements.

Part a(ii)

Empty set and universal set.

  • An empty set (or null set) is a set that contains no elements. It is denoted by ∅\emptyset or {}.
  • A universal set is a set that contains all elements relevant to a particular context or problem. It is usually denoted by UU.

Part b(i)

35.939\text{35.939}

Harmonic Mean (HM)

Calculate the harmonic mean, mode, and median for the given data.

First, we prepare a table with midpoints (xx), frequencies (ff), 1/x1/x, f/xf/x, and cumulative frequencies (cfcf).

MarksNumber of Students (ff)Midpoint (xx)1/x1/xf/xf/xCumulative Frequency (cfcf)
0 - 10250.20.42
10 - 207150.0666670.4666699
20 - 3015250.040.624
30 - 4025350.0285710.71427549
40 - 5045450.0222221.094
50 - 6032550.0181820.581824126
60 - 7015650.0153850.230775141
70 - 809750.0133330.12150
Total∑f=N=150\sum f = N = 150∑(f/x)≈4.173543\sum (f/x) \approx 4.173543

The formula for the harmonic mean for grouped data is:

HM=N∑fxHM = \frac{N}{\sum \frac{f}{x}}
  1. Sum the frequencies (NN) and the values of f/xf/x.

    N=150N = 150
    ∑fx=0.4+0.466669+0.6+0.714275+1.0+0.581824+0.230775+0.12≈4.173543\sum \frac{f}{x} = 0.4 + 0.466669 + 0.6 + 0.714275 + 1.0 + 0.581824 + 0.230775 + 0.12 \approx 4.173543

  2. Calculate the harmonic mean.

    HM=1504.173543HM = \frac{150}{4.173543} HM≈35.939HM \approx 35.939

    The harmonic mean is 35.939\boxed{\text{35.939}}.

Part b(ii)

46.061\text{46.061}

Mode

The formula for the mode for grouped data is:

Mode=L+(fm−f12fm−f1−f2)×hMode = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h

The modal class is 40 - 50 (highest frequency, fm=45f_m = 45).
L=40L = 40 (lower boundary of the modal class)
fm=45f_m = 45 (frequency of the modal class)
f1=25f_1 = 25 (frequency of the preceding class, 30 - 40)
f2=32f_2 = 32 (frequency of the succeeding class, 50 - 60)
h=10h = 10 (class width)

  1. Substitute the values into the formula.

    Mode=40+(45−252(45)−25−32)×10Mode = 40 + \left( \frac{45 - 25}{2(45) - 25 - 32} \right) \times 10
  2. Simplify the expression.

    Mode=40+(2090−57)×10Mode = 40 + \left( \frac{20}{90 - 57} \right) \times 10 Mode=40+(2033)×10Mode = 40 + \left( \frac{20}{33} \right) \times 10 Mode=40+0.60606×10Mode = 40 + 0.60606 \times 10 Mode=40+6.0606Mode = 40 + 6.0606 Mode≈46.061Mode \approx 46.061

    The mode is 46.061\boxed{\text{46.061}}.

Part b(iii)

45.778\text{45.778}

Median

The formula for the median for grouped data is:

Median=L+(N2−cfbfm)×hMedian = L + \left( \frac{\frac{N}{2} - cf_b}{f_m} \right) \times h

First, find the median position: N2=1502=75\frac{N}{2} = \frac{150}{2} = 75.
The median class is 40 - 50 (where the 75th observation falls, as cfcf goes from 49 to 94).
L=40L = 40 (lower boundary of the median class)
N=150N = 150 (total frequency)
cfb=49cf_b = 49 (cumulative frequency of the class before the median class)
fm=45f_m = 45 (frequency of the median class)
h=10h = 10 (class width)

  1. Substitute the values into the formula.

    Median=40+(75−4945)×10Median = 40 + \left( \frac{75 - 49}{45} \right) \times 10
  2. Simplify the expression.

    Median=40+(2645)×10Median = 40 + \left( \frac{26}{45} \right) \times 10 Median=40+0.57777×10Median = 40 + 0.57777 \times 10 Median=40+5.7777Median = 40 + 5.7777 Median≈45.778Median \approx 45.778

    The median is 45.778\boxed{\text{45.778}}.

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