Goldentiger44idn
Answered

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100 people answered the question "What is the area of your garden?"

There were twice as many responses in the interval [tex]\(20 \ \textless \ x \leq 30\)[/tex] as there were in the interval [tex]\(30 \ \textless \ x \leq 40\)[/tex]. Work out an estimate of the mean garden area. Give your answer to 1 decimal place.

[tex]\[
\begin{tabular}{|c|c|}
\hline
Area, \(x\left(m^2\right)\) & Frequency \\
\hline
\(0 \ \textless \ x \leq 10\) & 16 \\
\hline
\(10 \ \textless \ x \leq 20\) & 44 \\
\hline
\(20 \ \textless \ x \leq 30\) & \(\square\) \\
\hline
\(30 \ \textless \ x \leq 40\) & \(\square\) \\
\hline
\(40 \ \textless \ x \leq 50\) & 4 \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
To work out an estimate of the mean garden area, let's go through the problem step-by-step.

1. Interpreting the information:
- We know the frequencies for the ranges [tex]\(0 < x \leq 10\)[/tex], [tex]\(10 < x \leq 20\)[/tex], and [tex]\(40 < x \leq 50\)[/tex].
- The frequencies are:
- [tex]\(0 < x \leq 10\)[/tex]: 16
- [tex]\(10 < x \leq 20\)[/tex]: 44
- [tex]\(40 < x \leq 50\)[/tex]: 4

2. Finding the frequencies for the unknown intervals:
- We are told that the number of responses in the interval [tex]\(20 < x \leq 30\)[/tex] is twice that in the interval [tex]\(30 < x \leq 40\)[/tex].
- Let the frequency for the interval [tex]\(30 < x \leq 40\)[/tex] be [tex]\(x\)[/tex].
- Then, the frequency for the interval [tex]\(20 < x \leq 30\)[/tex] will be [tex]\(2x\)[/tex].

3. Creating an equation for the total number of responses:
- The total number of responses is 100.
- Therefore, we can set up the equation:
[tex]\[ 16 + 44 + 2x + x + 4 = 100 \][/tex]
Simplify:
[tex]\[ 64 + 3x = 100 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 3x = 36 \implies x = 12 \][/tex]
- Thus, the frequency for the interval [tex]\(30 < x \leq 40\)[/tex] is 12, and for the interval [tex]\(20 < x \leq 30\)[/tex] is [tex]\(2 \times 12 = 24\)[/tex].

4. Calculating midpoints for each interval:
- For the interval [tex]\(0 < x \leq 10\)[/tex], midpoint = [tex]\(\frac{0 + 10}{2} = 5\)[/tex]
- For the interval [tex]\(10 < x \leq 20\)[/tex], midpoint = [tex]\(\frac{10 + 20}{2} = 15\)[/tex]
- For the interval [tex]\(20 < x \leq 30\)[/tex], midpoint = [tex]\(\frac{20 + 30}{2} = 25\)[/tex]
- For the interval [tex]\(30 < x \leq 40\)[/tex], midpoint = [tex]\(\frac{30 + 40}{2} = 35\)[/tex]
- For the interval [tex]\(40 < x \leq 50\)[/tex], midpoint = [tex]\(\frac{40 + 50}{2} = 45\)[/tex]

5. Calculating the total frequency-weighted area:
- Sum of all frequencies [tex]\(= 16 + 44 + 24 + 12 + 4 = 100\)[/tex]
- Total area calculated using midpoints and frequencies:
[tex]\[ (5 \times 16) + (15 \times 44) + (25 \times 24) + (35 \times 12) + (45 \times 4) = 80 + 660 + 600 + 420 + 180 = 1940 \][/tex]

6. Calculating the mean garden area:
- Mean [tex]\(= \frac{\text{total area}}{\text{total frequency}} = \frac{1940}{100} = 19.4\)[/tex]

Hence, the estimated mean garden area is [tex]\(\boxed{19.4 \ \text{m}^2}\)[/tex].
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