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The quartile divides the distribution into four equal parts. When n is large they are given as Lower quartile Q1 = \( \frac{n}{4} \scriptsize th \; value \)

Middle Quartile or Median Q2 = \( \frac{n}{2} \scriptsize th \; value \)

Upper Quartile Q3 = \( \frac{3n}{4} \scriptsize th \; value \)

However, when n ≤ 50 (i.e.discrete)

Q1 = \( \frac{n+1}{4} \scriptsize th \; value \)

Q2 = \( \frac{n+1}{2} \scriptsize th \; value \)

Q3 = \( \frac{3(n+1)}{4} \scriptsize th \; value \)

INTERQUARTILE RANGE

Interquartile range is the difference between the upper quartile and the lower quartile i.e. Interquartile range = Q3 – Q1

SEMI-INTERQUARTILE RANGE

Semi-interquartile range can be found by Q= \( \frac{1}{2} \)(Q3 – Q1).

PERCENTILES

Definition: The percentile divides the data into 100 equal parts, a vertical axis is drawn on the right hand side of the cumulative frequency curve/orgive and marked 0% to 100%.

  • The lower quartile or 1st quartile is the 25th percentile
  • The median or the 2nd quartile is the 50th percentile
  • The upper quartile or 3rd quartile is the 75th percentile

Example 1

The record of how eggs were laid by chickens in a poultry farm in a year is as given below.

Number of Eggs per YearNumber of Chickens
45 – 4910
50 – 5436
55 – 5964
60 – 6452
65 – 6928
70 – 7410

(i)Draw a cumulative frequency curve of the distribution

(ii) Use your graph to find the semi-interquartile range

(iii) Using a percentage axis on the right of the ogive, find the number of eggs the 70th percentile Chicken laid.

(iv) If a customer buys a chicken from the farm, what is the probability that the chicken lays at least 60eggs in a year? (King’s College)

CUMULATIVE FREQUENCY DIAGRAM 

Screen Shot 2020 10 14 at 11.06.46 PM
No. Of Eggs per YearFrequency(f)Cumulative Frequency(cf)Class Boundaries
45 – 49101044.5 – 49.5
50 – 54364649.5 – 54.5
55 – 596411054.5 – 59.5
60 – 645216259.5 – 64.5
65 – 692819064.5 – 69.5
70 – 741020069.5 – 74.5

(ii)

Q1 = \( \frac{200}{4}\scriptsize = 50 \)   → 54.5  (See graph)

Q3 = \( \frac{200 \; \times \; 3}{4}\scriptsize = 150 \)           → 62.5 (see graph)  

Semi – interquartile range = \( \frac{Q_3 \; – \; Q_1}{2} \)

= \( \frac{62.5 \; – \; 54.5}{2} = \frac{8}{2}\)

= 4.

(iii) 70th percentile Chicken laid 60 eggs (See graph)

(iv) At least 60 eggs  → we have 52 + 28 + 10

i.e. at least 60 eggs ≡ 90 Chickens

Total Number  ≡ 200 Chickens∴  Probability at least 60 eggs

= \( \frac{90}{200} = \frac{9}{20} \scriptsize = 0.45 \)

Example 2

The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.

Marks (%)44–4647-4950-5253-5556-5859-6162-6465-6768-7071-73
Frequency2511202642463693

(i) Draw a cumulative frequency curve of the distribution.

(ii) Use your graph to estimate the median mark

(iii) Use algebraic means to find the mode  and median mark

(iv) What percentage of the candidates will be selected if the cut-off mark was 60%?

(v)Calculate the mean mark. (King’s College, Lagos).

Marks (%)Frequency(f)Cumulative Frequency(cf)Class BoundariesMid-Class (x)fx
44 – 462243.5 – 46.54590
47 – 495746.5 – 49.548240
50 – 52111849.5 – 52.551561
53 – 55203852.5 – 55.5541080
56 – 58266455.5 – 58.5571482
59 – 614210658.5 – 61.5602520
62 – 644615261.5 – 64.5632898
65 – 673618864.5 – 67.5662376
68 – 70919767.5 – 70.569621
71 – 73320070.5 – 73.572216
Total20012,084

(ii) Median Mark = 61 (See graph)

(iii) Mode = \( \scriptsize L + \frac {C \left (fm \;- \;fb \right)}{\left (fm \;- \; fb \right) + \left (fm \;- \; fa \right)} \)

Therefore, Mode = 61.5 + \( \frac {3 \left (46 \;- \; 42 \right)}{\left (46 \;- \; 42 \right) + \left (46 \;- \; 36 \right)} \)

= \( \scriptsize 61.5 + \normalsize \frac {3(4)}{4 + 10} \)

= \( \scriptsize 61.5 + \normalsize \frac {12}{14} \)

= 61.5 + 0.85714286

= 62.3571429

Therefore, Mode = 62.

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