Math  /  Data & Statistics

Question(a) Kenworth Electrical specialises in wiring new houses.
The monthly wages of all Kenworth Electrical employees are summarised in the frequency table below. \begin{tabular}{|c|c|} \hline Monthly wage, £x£ x & Frequency \\ \hline 1800x<20001800 \leqslant x<2000 & 64 \\ \hline 2000x<21002000 \leqslant x<2100 & 50 \\ \hline 2100x<24002100 \leqslant x<2400 & 2 \\ \hline 2400x<58002400 \leqslant x<5800 & 0 \\ \hline 5800x<78005800 \leqslant x<7800 & 4 \\ \hline \end{tabular} i) Estimate the mean ii) Which group contains the median iii) Write down the modal group

Studdy Solution

STEP 1

What is this asking? We need to find the average monthly wage, the wage group where the middle earner falls, and the most common wage group at Kenworth Electrical. Watch out! Don't forget that the last group is much wider than the others!

STEP 2

1. Estimate the Mean
2. Find the Median Group
3. Find the Modal Group

STEP 3

Alright, let's **estimate the mean**!
To do this with grouped data, we pretend everyone in each group earns the midpoint of that group's wage range.

STEP 4

For the first group, 1800x<20001800 \leqslant x < 2000, the midpoint is 1800+20002=1900\frac{1800 + 2000}{2} = 1900.
There are **64** people in this group.

STEP 5

For the second group, 2000x<21002000 \leqslant x < 2100, the midpoint is 2000+21002=2050\frac{2000 + 2100}{2} = 2050.
There are **50** people earning around this much.

STEP 6

Next, 2100x<24002100 \leqslant x < 2400 has a midpoint of 2100+24002=2250\frac{2100 + 2400}{2} = 2250 with **2** people.

STEP 7

No one earns between 24002400 and 58005800, so we can skip this group!

STEP 8

Finally, for 5800x<78005800 \leqslant x < 7800, the midpoint is 5800+78002=6800\frac{5800 + 7800}{2} = 6800 with **4** big earners!

STEP 9

Now, let's find the total estimated wages.
We multiply each midpoint by the number of people in that group and add it all up: (190064)+(205050)+(22502)+(68004)=121600+102500+4500+27200=256000(1900 \cdot 64) + (2050 \cdot 50) + (2250 \cdot 2) + (6800 \cdot 4) = 121600 + 102500 + 4500 + 27200 = 256000.

STEP 10

The total number of employees is 64+50+2+4=12064 + 50 + 2 + 4 = 120.

STEP 11

So, our **estimated mean** monthly wage is 2558001202131.67\frac{255800}{120} \approx 2131.67.

STEP 12

To find the **median group**, we need to find where the middle earner falls.
Since there are **120** employees, the middle person is the 1202=60\frac{120}{2} = 60th person when the data is ordered.

STEP 13

The first group has **64** people, so the **60th person** falls within the first group, 1800x<20001800 \leqslant x < 2000.

STEP 14

The **modal group** is the group with the most people in it.
That's the first group, 1800x<20001800 \leqslant x < 2000, with **64** employees.

STEP 15

i) The estimated mean monthly wage is $2131.67\$2131.67. ii) The median falls in the 1800x<20001800 \leqslant x < 2000 group. iii) The modal group is 1800x<20001800 \leqslant x < 2000.

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