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Sagot :
Let's solve the problem step-by-step.
### Part (a): Estimating the Mode
1. Identify the modal class:
- The modal class is the class interval with the highest frequency.
- From the frequency distribution, we see that the highest frequency is [tex]\(11\)[/tex], which corresponds to the length interval [tex]\(175-179\)[/tex].
2. Determine the mode using the given formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_{m-1}}{(f_m - f_{m-1}) + (f_m - f_{m+1})}\right) \times h \][/tex]
where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class.
- [tex]\(f_m\)[/tex] is the frequency of the modal class.
- [tex]\(f_{m-1}\)[/tex] is the frequency of the class preceding the modal class.
- [tex]\(f_{m+1}\)[/tex] is the frequency of the class succeeding the modal class.
- [tex]\(h\)[/tex] is the class width.
Now, let's plug in the values:
- The lower class boundary [tex]\(L\)[/tex] of the modal interval [tex]\(175-179\)[/tex] is [tex]\(175 - 0.5 = 174.5\)[/tex].
- [tex]\(f_m = 11\)[/tex]
- [tex]\(f_{m-1} = 9\)[/tex] (frequency of the class [tex]\(170-174\)[/tex])
- [tex]\(f_{m+1} = 6\)[/tex] (frequency of the class [tex]\(180-184\)[/tex])
- The class width [tex]\(h\)[/tex] is 5 (since each interval spans 5 units: [tex]\(154 - 150 + 1 = 5\)[/tex], etc.)
[tex]\[ \text{Mode} = 174.5 + \left(\frac{11 - 9}{(11 - 9) + (11 - 6)}\right) \times 5 \][/tex]
3. Calculate the mode:
[tex]\[ \text{Mode} = 174.5 + \left(\frac{2}{2 + 5}\right) \times 5 = 174.5 + \left(\frac{2}{7}\right) \times 5 = 174.5 + 1.4286 \approx 175.93 \][/tex]
### Part (b): Calculating the Mean
1. Calculate the midpoints of each class:
- Midpoint of [tex]\(150-154\)[/tex] is [tex]\(\frac{150 + 154}{2} = 152\)[/tex]
- Midpoint of [tex]\(155-159\)[/tex] is [tex]\(\frac{155 + 159}{2} = 157\)[/tex]
- Midpoint of [tex]\(160-164\)[/tex] is [tex]\(\frac{160 + 164}{2} = 162\)[/tex]
- Midpoint of [tex]\(165-169\)[/tex] is [tex]\(\frac{165 + 169}{2} = 167\)[/tex]
- Midpoint of [tex]\(170-174\)[/tex] is [tex]\(\frac{170 + 174}{2} = 172\)[/tex]
- Midpoint of [tex]\(175-179\)[/tex] is [tex]\(\frac{175 + 179}{2} = 177\)[/tex]
- Midpoint of [tex]\(180-184\)[/tex] is [tex]\(\frac{180 + 184}{2} = 182\)[/tex]
- Midpoint of [tex]\(185-189\)[/tex] is [tex]\(\frac{185 + 189}{2} = 187\)[/tex]
2. Calculate the mean using the midpoints and frequencies:
[tex]\[ \text{Mean} = \frac{\sum (f \times x)}{\sum f} \][/tex]
where:
- [tex]\(f\)[/tex] is the frequency of each interval.
- [tex]\(x\)[/tex] is the midpoint of each interval.
Calculate [tex]\(\sum (f \times x)\)[/tex]:
[tex]\[ \begin{align*} \sum (f \times x) &= 5 \times 152 + 2 \times 157 + 6 \times 162 + 8 \times 167 + 9 \times 172 + 11 \times 177 + 6 \times 182 + 3 \times 187 \\ &= 760 + 314 + 972 + 1336 + 1548 + 1947 + 1092 + 561 \\ &= 8530 \end{align*} \][/tex]
Calculate [tex]\(\sum f\)[/tex]:
[tex]\[ \sum f = 5 + 2 + 6 + 8 + 9 + 11 + 6 + 3 = 50 \][/tex]
Now, calculate the mean:
[tex]\[ \text{Mean} = \frac{8530}{50} = 170.6 \][/tex]
### Final Answers:
(a) The mode of the distribution is approximately [tex]\(175.93\)[/tex] [tex]\(mm\)[/tex].
(b) The mean of the distribution is [tex]\(170.6\)[/tex] [tex]\(mm\)[/tex].
### Part (a): Estimating the Mode
1. Identify the modal class:
- The modal class is the class interval with the highest frequency.
- From the frequency distribution, we see that the highest frequency is [tex]\(11\)[/tex], which corresponds to the length interval [tex]\(175-179\)[/tex].
2. Determine the mode using the given formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_{m-1}}{(f_m - f_{m-1}) + (f_m - f_{m+1})}\right) \times h \][/tex]
where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class.
- [tex]\(f_m\)[/tex] is the frequency of the modal class.
- [tex]\(f_{m-1}\)[/tex] is the frequency of the class preceding the modal class.
- [tex]\(f_{m+1}\)[/tex] is the frequency of the class succeeding the modal class.
- [tex]\(h\)[/tex] is the class width.
Now, let's plug in the values:
- The lower class boundary [tex]\(L\)[/tex] of the modal interval [tex]\(175-179\)[/tex] is [tex]\(175 - 0.5 = 174.5\)[/tex].
- [tex]\(f_m = 11\)[/tex]
- [tex]\(f_{m-1} = 9\)[/tex] (frequency of the class [tex]\(170-174\)[/tex])
- [tex]\(f_{m+1} = 6\)[/tex] (frequency of the class [tex]\(180-184\)[/tex])
- The class width [tex]\(h\)[/tex] is 5 (since each interval spans 5 units: [tex]\(154 - 150 + 1 = 5\)[/tex], etc.)
[tex]\[ \text{Mode} = 174.5 + \left(\frac{11 - 9}{(11 - 9) + (11 - 6)}\right) \times 5 \][/tex]
3. Calculate the mode:
[tex]\[ \text{Mode} = 174.5 + \left(\frac{2}{2 + 5}\right) \times 5 = 174.5 + \left(\frac{2}{7}\right) \times 5 = 174.5 + 1.4286 \approx 175.93 \][/tex]
### Part (b): Calculating the Mean
1. Calculate the midpoints of each class:
- Midpoint of [tex]\(150-154\)[/tex] is [tex]\(\frac{150 + 154}{2} = 152\)[/tex]
- Midpoint of [tex]\(155-159\)[/tex] is [tex]\(\frac{155 + 159}{2} = 157\)[/tex]
- Midpoint of [tex]\(160-164\)[/tex] is [tex]\(\frac{160 + 164}{2} = 162\)[/tex]
- Midpoint of [tex]\(165-169\)[/tex] is [tex]\(\frac{165 + 169}{2} = 167\)[/tex]
- Midpoint of [tex]\(170-174\)[/tex] is [tex]\(\frac{170 + 174}{2} = 172\)[/tex]
- Midpoint of [tex]\(175-179\)[/tex] is [tex]\(\frac{175 + 179}{2} = 177\)[/tex]
- Midpoint of [tex]\(180-184\)[/tex] is [tex]\(\frac{180 + 184}{2} = 182\)[/tex]
- Midpoint of [tex]\(185-189\)[/tex] is [tex]\(\frac{185 + 189}{2} = 187\)[/tex]
2. Calculate the mean using the midpoints and frequencies:
[tex]\[ \text{Mean} = \frac{\sum (f \times x)}{\sum f} \][/tex]
where:
- [tex]\(f\)[/tex] is the frequency of each interval.
- [tex]\(x\)[/tex] is the midpoint of each interval.
Calculate [tex]\(\sum (f \times x)\)[/tex]:
[tex]\[ \begin{align*} \sum (f \times x) &= 5 \times 152 + 2 \times 157 + 6 \times 162 + 8 \times 167 + 9 \times 172 + 11 \times 177 + 6 \times 182 + 3 \times 187 \\ &= 760 + 314 + 972 + 1336 + 1548 + 1947 + 1092 + 561 \\ &= 8530 \end{align*} \][/tex]
Calculate [tex]\(\sum f\)[/tex]:
[tex]\[ \sum f = 5 + 2 + 6 + 8 + 9 + 11 + 6 + 3 = 50 \][/tex]
Now, calculate the mean:
[tex]\[ \text{Mean} = \frac{8530}{50} = 170.6 \][/tex]
### Final Answers:
(a) The mode of the distribution is approximately [tex]\(175.93\)[/tex] [tex]\(mm\)[/tex].
(b) The mean of the distribution is [tex]\(170.6\)[/tex] [tex]\(mm\)[/tex].
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