1. A gardener digs up his carrots and measures their lengths (to the nearest mm) and groups the results as shown below:

| Length (mm) | 150-154 | 155-159 | 160-164 | 165-169 | 170-174 | 175-179 | 180-184 | 185-189 |
|-------------|---------|---------|---------|---------|---------|---------|---------|---------|
| Frequency | 5 | 2 | 6 | 8 | 9 | 11 | 6 | 3 |

(a) Estimate the mode of the distribution, correct to two decimal places.

(b) Calculate the mean of the distribution, correct to 4 significant figures.



Answer :

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].