The frequency distribution table below shows the mass, in grams, of 50 industrial steel be… — JAMB UTME

The frequency distribution table below shows the mass, in grams, of $50$ industrial steel bearings measured during a precision manufacturing audit:

If the mean mass of the bearings is $34.7\text{ g}$ , what is the value of the missing frequency $x$ ?

Answer: 18

The value of

x

is 18 because setting up the total frequency sum gives

x + y = 25

, and using class midpoints to compute the mean yields

1915 - 10x = 1735

, which solves to

x = 18

Express the relationship between the missing frequencies using total frequency.

x + y = 25

y = 25 - x

The total number of industrial steel bearings is 50, so

5 + 12 + x + y + 8 = 50

Determine the midpoint (

m

) of each class interval.

Midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5 respectively.

The class midpoint is calculated as

\frac{\text{lower limit} + \text{upper limit}}{2}

Formulate the equation for the sum of products of frequencies and midpoints.

\sum fm = 802.5 + 34.5x + 44.5y

Multiply each class midpoint by its corresponding frequency and sum the results.

Substitute

y = 25 - x

and solve for

x

using the mean formula.

x = 18

Setting

\frac{1915 - 10x}{50} = 34.7

yields

1915 - 10x = 1735

, which gives

10x = 180

and thus

x = 18

Measures of Central Tendency for Grouped Data - Mean with Unknown Frequencies

Estimated Time:3m 0s

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