A survey of workers' wages in a small company found the following distribution. If the mean wage was 27…

Question

A survey of workers' wages in a small company found the following distribution. If the mean wage was 27 rac1116, find the value of a.

ScanSolve AI Tutor · Accepted Answer

Here's the solution to question 17: Part (a): Find the value of a. Step 1: Calculate the midpoint (x) for each class interval and the product of frequency and midpoint (fx). The mean of grouped data is given by the formula: Mean = (∑ fx)/(∑ f). Amount (USD) — Number of workers (f) — Midpoint (x) — fx10 - 14 — 6 — (10+14)/(2) = 12 — 6 × 12 = 72; 15 − 24 — 16 — (15+24)/(2) = 19.5 — 16 × 19.5 = 312; 25 − 29 — a — (25+29)/(2) = 27 — a × 27 = 27a; 30 − 44 — 10 — (30+44)/(2) = 37 — 10 × 37 = 370; 45 − 64 — 5 — (45+64)/(2) = 54.5 — 5 × 54.5 = 272.5 Step 2: Calculate the sum of frequencies (∑ f) and the sum of fx (∑ fx). ∑ f = 6 + 16 + a + 10 + 5 = 37 + a; ∑ fx = 72 + 312 + 27a + 370 + 272.5 = 1026.5 + 27a Step 3: Use the given mean wage to set up an equation. The mean wage is 27(11)/(16) USD. Convert this to an improper fraction: 27(11)/(16) = (27 × 16 + 11)/(16) = (432 + 11)/(16) = 443/16 Now, apply the mean formula: (1026.5 + 27a)/(37 + a) = 443/16 Step 4: Solve for a. Cross-multiply: 16(1026.5 + 27a) = 443(37 + a); 16424 + 432a = 16391 + 443a Rearrange the terms to isolate a: 16424 − 16391 = 443a − 432a; 33 = 11a; a = 33/11; a = 3 The value of a is 3. Part (b): Find the median class and its frequency. Step 1: Calculate the total number of workers (∑ f). Using a=3 from part (a): ∑ f = 37 + a = 37 + 3 = 40 Step 2: Determine the position of the median. The median position is (∑ f)/(2) = 40/2 = 20^(th) position. Step 3: Find the median class using cumulative frequency. Amount (USD) — Number of workers (f) — Cumulative Frequency (CF)10 - 14 — 6 — 6 15 − 24 — 16 — 6 + 16 = 22; 25 − 29 — 3 — 22 + 3 = 25; 30 − 44 — 10 — 25 + 10 = 35; 45 − 64 — 5 — 35 + 5 = 40 The 20^(th) worker falls into the class where the cumulative frequency first exceeds 20. This is the 15 − 24 class, which has a cumulative frequency of 22. The median class is 15 − 24. The frequency of the median class is 16. Part (c): On the grid provided, draw a histogram to represent the information. To draw a histogram for data with unequal class widths, we need to calculate the frequency density for each class. Frequency Density = FrequencyClass Width Step 1: Determine class boundaries and class widths. Amount (USD) — Number of workers (f) — Class Boundaries — Class Width — Frequency Density10 - 14 — 6 — 9.5 − 14.5 — 14.5 − 9.5 = 5 — 6/5 = 1.2; 15 − 24 — 16 — 14.5 − 24.5 — 24.5 − 14.5 = 10 — 16/10 = 1.6; 25 − 29 — 3 — 24.5 − 29.5 — 29.5 − 24.5 = 5 — 3/5 = 0.6; 30 − 44 — 10 — 29.5 − 44.5 — 44.5 − 29.5 = 15 — 10/15 ≈ 0.67; 45 − 64 — 5 — 44.5 − 64.5 — 64.5 − 44.5 = 20 — 5/20 = 0.25 Step 2: Draw the histogram. X-axis: Label "Amount (USD)". Mark the class boundaries: 9.5, 14.5, 24.5, 29.5, 44.5, 64.5. Y-axis: Label "Frequency Density". Choose a suitable scale, for example, 1 unit on the grid represents 0.2 frequency density, going up to at least 1.8. Draw rectangles for each class: 1. 10 − 14 (9.5 − 14.5): Draw a bar from 9.5 to 14.5 on the x-axis, with a height of 1.2 on the y-axis. 2. 15 − 24 (14.5 − 24.5): Draw a bar from 14.5 to 24.5 on the x-axis, with a height of 1.6 on the y-axis. 3. **25 -

Amount (USD)	Number of workers ( $f$ )	Midpoint ( $x$ )	$fx$
10 - 14	6	$\frac{10+14}{2} = 12$	$6 \times 12 = 72$
15 - 24	16	$\frac{15+24}{2} = 19.5$	$16 \times 19.5 = 312$
25 - 29	$a$	$\frac{25+29}{2} = 27$	$a \times 27 = 27a$
30 - 44	10	$\frac{30+44}{2} = 37$	$10 \times 37 = 370$
45 - 64	5	$\frac{45+64}{2} = 54.5$	$5 \times 54.5 = 272.5$

Amount (USD)	Number of workers ( $f$ )	Cumulative Frequency (CF)
10 - 14	6	6
15 - 24	16	$6 + 16 = 22$
25 - 29	3	$22 + 3 = 25$
30 - 44	10	$25 + 10 = 35$
45 - 64	5	$35 + 5 = 40$

Amount (USD)	Number of workers ( $f$ )	Class Boundaries	Class Width	Frequency Density
10 - 14	6	9.5 - 14.5	$14.5 - 9.5 = 5$	$\frac{6}{5} = 1.2$
15 - 24	16	14.5 - 24.5	$24.5 - 14.5 = 10$	$\frac{16}{10} = 1.6$
25 - 29	3	24.5 - 29.5	$29.5 - 24.5 = 5$	$\frac{3}{5} = 0.6$
30 - 44	10	29.5 - 44.5	$44.5 - 29.5 = 15$	$\frac{10}{15} \approx 0.67$
45 - 64	5	44.5 - 64.5	$64.5 - 44.5 = 20$	$\frac{5}{20} = 0.25$

A survey of workers' wages in a small company found the following distribution. If the mean wage was 27 rac1116, find the value of a.

Answers

Step-by-step solution

Part a· 4 steps

Calculate the midpoint ( $x$ ) for each class interval and the product of frequency and midpoint ( $fx$ ).

Calculate the sum of frequencies ( $\sum f$ ) and the sum of $fx$ ( $\sum fx$ ).

Use the given mean wage to set up an equation.

Solve for $a$ .

Part b· 3 steps
$\text{15 - 24}$

Calculate the total number of workers ( $\sum f$ ).

Determine the position of the median.

Find the median class using cumulative frequency.

Part c· 2 steps

Determine class boundaries and class widths.

Draw the histogram.

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More Mathematics Questions

A survey of workers' wages in a small company found the following distribution. If the mean wage was 27 rac1116, find the value of a.

Answers

Step-by-step solution

Part a· 4 steps

Calculate the midpoint (xxx) for each class interval and the product of frequency and midpoint (fxfxfx).

Calculate the sum of frequencies (∑f\sum f∑f) and the sum of fxfxfx (∑fx\sum fx∑fx).

Use the given mean wage to set up an equation.

Solve for aaa.

Part b· 3 steps15 - 24\text{15 - 24}15 - 24

Calculate the total number of workers (∑f\sum f∑f).

Determine the position of the median.

Find the median class using cumulative frequency.

Part c· 2 steps

Determine class boundaries and class widths.

Draw the histogram.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Calculate the midpoint ( $x$ ) for each class interval and the product of frequency and midpoint ( $fx$ ).

Calculate the sum of frequencies ( $\sum f$ ) and the sum of $fx$ ( $\sum fx$ ).

Solve for $a$ .

Part b· 3 steps
$\text{15 - 24}$

Calculate the total number of workers ( $\sum f$ ).