The table below shows the cumulative frequency distribution of the masses (in grams) of 100 cocoa beans sampled from an agricultural yield: | Mass Class Interval (g) | Cumulative Frequency | |---|---| | 10 - 19 | 10 | | 20 - 29 | 30 | | 30 - 39 | 65 | | 40 - 49 | 90 | | 50 - 59 | 100 | Using linear

The table below shows the cumulative frequency distribution of the masses (in grams) of 10… — JAMB UTME

The table below shows the cumulative frequency distribution of the masses (in grams) of $100$ cocoa beans sampled from an agricultural yield:

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile ( $Q_3$ ) mass of the cocoa beans in grams.

Answer: 43.5 g

The 75th percentile (

Q_3

) mass of the cocoa beans is

43.5\text{ g}

To find the 75th percentile (

Q_3

) from cumulative frequency data, calculate the rank

\frac{75}{100} \times 100 = 75

. This falls into the

40 - 49

class interval (boundaries

39.5 - 49.5

). Applying

Q_3 = L + \left(\frac{75 - F}{f}\right)c

yields

39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}

Calculate the percentile position rank

Rank position is

75

The 75th percentile corresponds to

75\%

of the total sample size

N = 100

, giving

\frac{75}{100} \times 100 = 75

Locate the 75th percentile class interval and its boundaries

Class interval is

40 - 49

, with lower boundary

L = 39.5

and upper boundary

49.5

Cumulative frequency before

40 - 49

65

, and up to

40 - 49

90

. Since

65 < 75 \leq 90

, the 75th item falls in this interval.

Identify class parameters for interpolation

L = 39.5

F = 65

f = 25

c = 10

Lower boundary

L = 39.5

, previous cumulative frequency

F = 65

, class frequency

f = 90 - 65 = 25

, class width

c = 49.5 - 39.5 = 10

Compute

Q_3

using the linear interpolation formula

Q_3 = 43.5\text{ g}

Q_3 = 39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 39.5 + 4 = 43.5

Linear Interpolation of Percentiles from Cumulative Frequency Data