The table below shows the distribution of masses (in kg) of cocoa bags harvested on a farm: | Mass (kg) | Frequency (f) | |---|---| | 10 - 19 | 6 | | 20 - 29 | k | | 30 - 39 | 15 | | 40 - 49 | 12 | | 50 - 59 | 7 | If the estimated mean mass of the distribution is 35.3\text{ kg}, calculate the value

The table below shows the distribution of masses (in kg) of cocoa bags harvested on a farm… — JAMB UTME

The table below shows the distribution of masses (in kg) of cocoa bags harvested on a farm:

If the estimated mean mass of the distribution is $35.3\text{ kg}$ , calculate the value of the missing frequency $k$ .

Answer: 10

The value of the missing frequency is 10.

To find the missing frequency, compute the midpoints (

x

) of each mass class interval: 14.5, 24.5, 34.5, 44.5, and 54.5. Next, express the sum of frequencies as

\sum f = 40 + k

and the sum of products of frequency and midpoint as

\sum fx = 6(14.5) + k(24.5) + 15(34.5) + 12(44.5) + 7(54.5) = 1520 + 24.5k

. Equating the mean expression

\frac{\sum fx}{\sum f}

35.3

gives

\frac{1520 + 24.5k}{40 + k} = 35.3

. Cross-multiplying and solving yields

1520 + 24.5k = 1412 + 35.3k

, which simplifies to

10.8k = 108

, giving

k = 10

Determine the class midpoints (

x

) for all intervals.

Class midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5.

Grouped mean calculations require representative midpoint values for each class interval.

Formulate expressions for total frequency

\sum f

and total weighted sum

\sum fx

\sum f = 40 + k and \sum fx = 1520 + 24.5k.

These algebraic expressions are necessary to substitute into the mean formula.

Set up and solve the linear equation using the given mean of 35.3.

\frac{1520 + 24.5k}{40 + k} = 35.3 \implies 10.8k = 108 \implies k = 10.

Equating the algebraic mean expression to the numerical mean allows solving for the unknown frequency k.

Measures of Central Tendency for Grouped Data