Statistics

Given: Data: 4, 7, 8, 9, 10, 12, 13, 17, $n = 8$

Step 1: Find the Mean
$$\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$$

Step 2: Find $|x_i - \bar{x}|$ for each observation

| $x_i$ | $|x_i - \bar{x}|$ |
|---|---|
| 4 | 6 |
| 7 | 3 |
| 8 | 2 |
| 9 | 1 |
| 10 | 0 |
| 12 | 2 |
| 13 | 3 |
| 17 | 7 |
| Total | 24 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum|x_i - \bar{x}|}{n} = \frac{24}{8} = 3$$

Answer: Mean Deviation about the mean $= 3$

Given: Data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44, $n = 10$

Step 1: Find the Mean
$$\bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} = 50$$

Step 2: Find $|x_i - \bar{x}|$ for each observation

| $x_i$ | $|x_i - 50|$ |
|---|---|
| 38 | 12 |
| 70 | 20 |
| 48 | 2 |
| 40 | 10 |
| 42 | 8 |
| 55 | 5 |
| 63 | 13 |
| 46 | 4 |
| 54 | 4 |
| 44 | 6 |
| Total | 84 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum|x_i - \bar{x}|}{n} = \frac{84}{10} = 8.4$$

Answer: Mean Deviation about the mean $= 8.4$

Given: Data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17, $n = 12$

Step 1: Arrange in ascending order
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Step 2: Find the Median
Since $n = 12$ (even),
$$\text{Median} = \frac{\text{6th term} + \text{7th term}}{2} = \frac{13 + 14}{2} = 13.5$$

Step 3: Find $|x_i - M|$ for each observation

| $x_i$ | $|x_i - 13.5|$ |
|---|---|
| 10 | 3.5 |
| 11 | 2.5 |
| 11 | 2.5 |
| 12 | 1.5 |
| 13 | 0.5 |
| 13 | 0.5 |
| 14 | 0.5 |
| 16 | 2.5 |
| 16 | 2.5 |
| 17 | 3.5 |
| 17 | 3.5 |
| 18 | 4.5 |
| Total | 28 |

Step 4: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum|x_i - M|}{n} = \frac{28}{12} = 2.33$$

Answer: Mean Deviation about the median $\approx 2.33$

Given: Data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49, $n = 10$

Step 1: Arrange in ascending order
36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Step 2: Find the Median
Since $n = 10$ (even),
$$\text{Median} = \frac{\text{5th term} + \text{6th term}}{2} = \frac{46 + 49}{2} = 47.5$$

Step 3: Find $|x_i - M|$ for each observation

| $x_i$ | $|x_i - 47.5|$ |
|---|---|
| 36 | 11.5 |
| 42 | 5.5 |
| 45 | 2.5 |
| 46 | 1.5 |
| 46 | 1.5 |
| 49 | 1.5 |
| 51 | 3.5 |
| 53 | 5.5 |
| 60 | 12.5 |
| 72 | 24.5 |
| Total | 70 |

Step 4: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum|x_i - M|}{n} = \frac{70}{10} = 7$$

Answer: Mean Deviation about the median $= 7$

Given: Discrete frequency distribution with $N = \sum f_i = 7+4+6+3+5 = 25$

Step 1: Find the Mean

| $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|
| 5 | 7 | 35 |
| 10 | 4 | 40 |
| 15 | 6 | 90 |
| 20 | 3 | 60 |
| 25 | 5 | 125 |
| Total | 25 | 350 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$$

Step 2: Find $f_i|x_i - \bar{x}|$

| $x_i$ | $f_i$ | $|x_i - 14|$ | $f_i|x_i - 14|$ |
|---|---|---|---|
| 5 | 7 | 9 | 63 |
| 10 | 4 | 4 | 16 |
| 15 | 6 | 1 | 6 |
| 20 | 3 | 6 | 18 |
| 25 | 5 | 11 | 55 |
| Total | 25 | | 158 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$$

Answer: Mean Deviation about the mean $= 6.32$

Given: Discrete frequency distribution with $N = \sum f_i = 4+24+28+16+8 = 80$

Step 1: Find the Mean

| $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|
| 10 | 4 | 40 |
| 30 | 24 | 720 |
| 50 | 28 | 1400 |
| 70 | 16 | 1120 |
| 90 | 8 | 720 |
| Total | 80 | 4000 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4000}{80} = 50$$

Step 2: Find $f_i|x_i - \bar{x}|$

| $x_i$ | $f_i$ | $|x_i - 50|$ | $f_i|x_i - 50|$ |
|---|---|---|---|
| 10 | 4 | 40 | 160 |
| 30 | 24 | 20 | 480 |
| 50 | 28 | 0 | 0 |
| 70 | 16 | 20 | 320 |
| 90 | 8 | 40 | 320 |
| Total | 80 | | 1280 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{1280}{80} = 16$$

Answer: Mean Deviation about the mean $= 16$

Given: $N = \sum f_i = 8+6+2+2+2+6 = 26$

Step 1: Find the Median

Compute cumulative frequencies:

| $x_i$ | $f_i$ | Cumulative Frequency |
|---|---|---|
| 5 | 8 | 8 |
| 7 | 6 | 14 |
| 9 | 2 | 16 |
| 10 | 2 | 18 |
| 12 | 2 | 20 |
| 15 | 6 | 26 |

$N = 26$, so $\frac{N}{2} = 13$.

The 13th and 14th observations both lie in the $x_i = 7$ group (cumulative frequency reaches 14 at $x_i = 7$).

$$\text{Median} = \frac{7+7}{2} = 7$$

Step 2: Find $f_i|x_i - M|$

| $x_i$ | $f_i$ | $|x_i - 7|$ | $f_i|x_i - 7|$ |
|---|---|---|---|
| 5 | 8 | 2 | 16 |
| 7 | 6 | 0 | 0 |
| 9 | 2 | 2 | 4 |
| 10 | 2 | 3 | 6 |
| 12 | 2 | 5 | 10 |
| 15 | 6 | 8 | 48 |
| Total | 26 | | 84 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{84}{26} = \frac{42}{13} \approx 3.23$$

Answer: Mean Deviation about the median $\approx 3.23$

Given: $N = \sum f_i = 3+5+6+7+8 = 29$

Step 1: Find the Median

Compute cumulative frequencies:

| $x_i$ | $f_i$ | Cumulative Frequency |
|---|---|---|
| 15 | 3 | 3 |
| 21 | 5 | 8 |
| 27 | 6 | 14 |
| 30 | 7 | 21 |
| 35 | 8 | 29 |

$\frac{N+1}{2} = \frac{30}{2} = 15$th observation.

The 15th observation lies in the group where $x_i = 30$ (cumulative frequency 21 covers positions 15 to 21).

$$\text{Median} = 30$$

Step 2: Find $f_i|x_i - M|$

| $x_i$ | $f_i$ | $|x_i - 30|$ | $f_i|x_i - 30|$ |
|---|---|---|---|
| 15 | 3 | 15 | 45 |
| 21 | 5 | 9 | 45 |
| 27 | 6 | 3 | 18 |
| 30 | 7 | 0 | 0 |
| 35 | 8 | 5 | 40 |
| Total | 29 | | 148 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{148}{29} \approx 5.1$$

Answer: Mean Deviation about the median $\approx 5.1$

Given: Continuous frequency distribution. $N = 4+8+9+10+7+5+4+3 = 50$

Step 1: Find midpoints and compute mean

| Class | $x_i$ (mid) | $f_i$ | $f_i x_i$ |
|---|---|---|---|
| 0–100 | 50 | 4 | 200 |
| 100–200 | 150 | 8 | 1200 |
| 200–300 | 250 | 9 | 2250 |
| 300–400 | 350 | 10 | 3500 |
| 400–500 | 450 | 7 | 3150 |
| 500–600 | 550 | 5 | 2750 |
| 600–700 | 650 | 4 | 2600 |
| 700–800 | 750 | 3 | 2250 |
| Total | | 50 | 17900 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{17900}{50} = 358$$

Step 2: Find $f_i|x_i - \bar{x}|$

| $x_i$ | $f_i$ | $|x_i - 358|$ | $f_i|x_i - 358|$ |
|---|---|---|---|
| 50 | 4 | 308 | 1232 |
| 150 | 8 | 208 | 1664 |
| 250 | 9 | 108 | 972 |
| 350 | 10 | 8 | 80 |
| 450 | 7 | 92 | 644 |
| 550 | 5 | 192 | 960 |
| 650 | 4 | 292 | 1168 |
| 750 | 3 | 392 | 1176 |
| Total | 50 | | 7896 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{7896}{50} = 157.92$$

Answer: Mean Deviation about the mean $= 157.92$

Given: Continuous frequency distribution. $N = 9+13+26+30+12+10 = 100$

Step 1: Find midpoints and compute mean

| Class | $x_i$ (mid) | $f_i$ | $f_i x_i$ |
|---|---|---|---|
| 95–105 | 100 | 9 | 900 |
| 105–115 | 110 | 13 | 1430 |
| 115–125 | 120 | 26 | 3120 |
| 125–135 | 130 | 30 | 3900 |
| 135–145 | 140 | 12 | 1680 |
| 145–155 | 150 | 10 | 1500 |
| Total | | 100 | 12530 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{12530}{100} = 125.3$$

Step 2: Find $f_i|x_i - \bar{x}|$

| $x_i$ | $f_i$ | $|x_i - 125.3|$ | $f_i|x_i - 125.3|$ |
|---|---|---|---|
| 100 | 9 | 25.3 | 227.7 |
| 110 | 13 | 15.3 | 198.9 |
| 120 | 26 | 5.3 | 137.8 |
| 130 | 30 | 4.7 | 141.0 |
| 140 | 12 | 14.7 | 176.4 |
| 150 | 10 | 24.7 | 247.0 |
| Total | 100 | | 1128.8 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{1128.8}{100} = 11.288$$

Answer: Mean Deviation about the mean $\approx 11.28$

Given: Continuous frequency distribution. $N = 6+8+14+16+4+2 = 50$

Step 1: Find the Median

Compute cumulative frequencies:

| Class | $f_i$ | Cumulative Frequency |
|---|---|---|
| 0–10 | 6 | 6 |
| 10–20 | 8 | 14 |
| 20–30 | 14 | 28 |
| 30–40 | 16 | 44 |
| 40–50 | 4 | 48 |
| 50–60 | 2 | 50 |

$\frac{N}{2} = 25$. The cumulative frequency just exceeding 25 is 28, so the median class is 20–30.

Using the formula:
$$\text{Median} = l + \frac{\frac{N}{2} - C}{f} \times h$$

Here $l = 20$, $C = 14$, $f = 14$, $h = 10$:
$$M = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + 7.857 \approx 27.86$$

Step 2: Find midpoints and $f_i|x_i - M|$

| Class | $x_i$ | $f_i$ | $|x_i - 27.86|$ | $f_i|x_i - 27.86|$ |
|---|---|---|---|---|
| 0–10 | 5 | 6 | 22.86 | 137.16 |
| 10–20 | 15 | 8 | 12.86 | 102.88 |
| 20–30 | 25 | 14 | 2.86 | 40.04 |
| 30–40 | 35 | 16 | 7.14 | 114.24 |
| 40–50 | 45 | 4 | 17.14 | 68.56 |
| 50–60 | 55 | 2 | 27.14 | 54.28 |
| Total | | 50 | | 517.16 |

Step 3: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{517.16}{50} \approx 10.34$$

Answer: Mean Deviation about the median $\approx 10.34$

Given: $N = 5+6+12+14+26+12+16+9 = 100$

Step 1: Convert to continuous classes (subtract 0.5 from lower, add 0.5 to upper)

| Class (continuous) | $x_i$ (mid) | $f_i$ | Cumulative Frequency |
|---|---|---|---|
| 15.5–20.5 | 18 | 5 | 5 |
| 20.5–25.5 | 23 | 6 | 11 |
| 25.5–30.5 | 28 | 12 | 23 |
| 30.5–35.5 | 33 | 14 | 37 |
| 35.5–40.5 | 38 | 26 | 63 |
| 40.5–45.5 | 43 | 12 | 75 |
| 45.5–50.5 | 48 | 16 | 91 |
| 50.5–55.5 | 53 | 9 | 100 |

Step 2: Find the Median

$\frac{N}{2} = 50$. The cumulative frequency just exceeding 50 is 63, so the median class is 35.5–40.5.

$$M = l + \frac{\frac{N}{2} - C}{f} \times h = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{65}{26} = 35.5 + 2.5 = 38$$

Step 3: Find $f_i|x_i - M|$

| $x_i$ | $f_i$ | $|x_i - 38|$ | $f_i|x_i - 38|$ |
|---|---|---|---|
| 18 | 5 | 20 | 100 |
| 23 | 6 | 15 | 90 |
| 28 | 12 | 10 | 120 |
| 33 | 14 | 5 | 70 |
| 38 | 26 | 0 | 0 |
| 43 | 12 | 5 | 60 |
| 48 | 16 | 10 | 160 |
| 53 | 9 | 15 | 135 |
| Total | 100 | | 735 |

Step 4: Calculate Mean Deviation
$$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{735}{100} = 7.35$$

Answer: Mean Deviation about the median $= 7.35$ years

Given: Data: 6, 7, 10, 12, 13, 4, 8, 12, $n = 8$

Step 1: Find the Mean
$$\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$$

Step 2: Find $(x_i - \bar{x})^2$

| $x_i$ | $x_i - 9$ | $(x_i - 9)^2$ |
|---|---|---|
| 6 | $-3$ | 9 |
| 7 | $-2$ | 4 |
| 10 | 1 | 1 |
| 12 | 3 | 9 |
| 13 | 4 | 16 |
| 4 | $-5$ | 25 |
| 8 | $-1$ | 1 |
| 12 | 3 | 9 |
| Total | | 74 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25$$

Answer: Mean $= 9$, Variance $= 9.25$

Given: First $n$ natural numbers: $1, 2, 3, \ldots, n$

Step 1: Find the Mean
$$\bar{x} = \frac{1+2+3+\cdots+n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$$

Step 2: Find the Variance

We use the formula:
$$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$

$$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\frac{\sum x_i^2}{n} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$$

$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$$

$$= \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$$

$$= (n+1)\left[\frac{2n+1}{6} - \frac{n+1}{4}\right]$$

$$= (n+1)\left[\frac{2(2n+1) - 3(n+1)}{12}\right]$$

$$= (n+1)\left[\frac{4n+2-3n-3}{12}\right]$$

$$= (n+1) \cdot \frac{n-1}{12} = \frac{n^2-1}{12}$$

Answer: Mean $= \dfrac{n+1}{2}$, Variance $= \dfrac{n^2-1}{12}$

Given: First 10 multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. $n = 10$

Step 1: Find the Mean
$$\bar{x} = \frac{3+6+9+\cdots+30}{10} = \frac{3(1+2+\cdots+10)}{10} = \frac{3 \times 55}{10} = \frac{165}{10} = 16.5$$

Step 2: Find $(x_i - \bar{x})^2$

| $x_i$ | $x_i - 16.5$ | $(x_i - 16.5)^2$ |
|---|---|---|
| 3 | $-13.5$ | 182.25 |
| 6 | $-10.5$ | 110.25 |
| 9 | $-7.5$ | 56.25 |
| 12 | $-4.5$ | 20.25 |
| 15 | $-1.5$ | 2.25 |
| 18 | 1.5 | 2.25 |
| 21 | 4.5 | 20.25 |
| 24 | 7.5 | 56.25 |
| 27 | 10.5 | 110.25 |
| 30 | 13.5 | 182.25 |
| Total | | 742.5 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{742.5}{10} = 74.25$$

Answer: Mean $= 16.5$, Variance $= 74.25$

Given: $N = \sum f_i = 2+4+7+12+8+4+3 = 40$

Step 1: Find the Mean

| $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|
| 6 | 2 | 12 |
| 10 | 4 | 40 |
| 14 | 7 | 98 |
| 18 | 12 | 216 |
| 24 | 8 | 192 |
| 28 | 4 | 112 |
| 30 | 3 | 90 |
| Total | 40 | 760 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{760}{40} = 19$$

Step 2: Find $f_i(x_i - \bar{x})^2$

| $x_i$ | $f_i$ | $(x_i - 19)^2$ | $f_i(x_i-19)^2$ |
|---|---|---|---|
| 6 | 2 | 169 | 338 |
| 10 | 4 | 81 | 324 |
| 14 | 7 | 25 | 175 |
| 18 | 12 | 1 | 12 |
| 24 | 8 | 25 | 200 |
| 28 | 4 | 81 | 324 |
| 30 | 3 | 121 | 363 |
| Total | 40 | | 1736 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{1736}{40} = 43.4$$

Answer: Mean $= 19$, Variance $= 43.4$

Given: $N = \sum f_i = 3+2+3+2+6+3+3 = 22$

Step 1: Find the Mean

| $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|
| 92 | 3 | 276 |
| 93 | 2 | 186 |
| 97 | 3 | 291 |
| 98 | 2 | 196 |
| 102 | 6 | 612 |
| 104 | 3 | 312 |
| 109 | 3 | 327 |
| Total | 22 | 2200 |

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2200}{22} = 100$$

Step 2: Find $f_i(x_i - \bar{x})^2$

| $x_i$ | $f_i$ | $(x_i - 100)^2$ | $f_i(x_i-100)^2$ |
|---|---|---|---|
| 92 | 3 | 64 | 192 |
| 93 | 2 | 49 | 98 |
| 97 | 3 | 9 | 27 |
| 98 | 2 | 4 | 8 |
| 102 | 6 | 4 | 24 |
| 104 | 3 | 16 | 48 |
| 109 | 3 | 81 | 243 |
| Total | 22 | | 640 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{640}{22} = 29.09$$

Answer: Mean $= 100$, Variance $\approx 29.09$

Given: $N = \sum f_i = 2+1+12+29+25+12+10+4+5 = 100$

Let assumed mean $A = 64$, $h = 1$, so $y_i = \frac{x_i - 64}{1} = x_i - 64$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

| $x_i$ | $f_i$ | $y_i = x_i - 64$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|
| 60 | 2 | $-4$ | $-8$ | 32 |
| 61 | 1 | $-3$ | $-3$ | 9 |
| 62 | 12 | $-2$ | $-24$ | 48 |
| 63 | 29 | $-1$ | $-29$ | 29 |
| 64 | 25 | 0 | 0 | 0 |
| 65 | 12 | 1 | 12 | 12 |
| 66 | 10 | 2 | 20 | 40 |
| 67 | 4 | 3 | 12 | 36 |
| 68 | 5 | 4 | 20 | 80 |
| Total | 100 | | 0 | 286 |

Step 2: Find Mean
$$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 64 + \frac{0}{100} \times 1 = 64$$

Step 3: Find Variance and Standard Deviation
$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$
$$= \frac{1}{(100)^2}\left[100 \times 286 - 0^2\right] = \frac{28600}{10000} = 2.86$$

$$\sigma = \sqrt{2.86} \approx 1.69$$

Answer: Mean $= 64$, Standard Deviation $\approx 1.69$

Given: $N = 2+3+5+10+3+5+2 = 30$

Step 1: Find midpoints and compute mean

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|---|
| 0–30 | 15 | 2 | 30 |
| 30–60 | 45 | 3 | 135 |
| 60–90 | 75 | 5 | 375 |
| 90–120 | 105 | 10 | 1050 |
| 120–150 | 135 | 3 | 405 |
| 150–180 | 165 | 5 | 825 |
| 180–210 | 195 | 2 | 390 |
| Total | | 30 | 3210 |

$$\bar{x} = \frac{3210}{30} = 107$$

Step 2: Find $f_i(x_i - \bar{x})^2$

| $x_i$ | $f_i$ | $(x_i - 107)^2$ | $f_i(x_i - 107)^2$ |
|---|---|---|---|
| 15 | 2 | 8464 | 16928 |
| 45 | 3 | 3844 | 11532 |
| 75 | 5 | 1024 | 5120 |
| 105 | 10 | 4 | 40 |
| 135 | 3 | 784 | 2352 |
| 165 | 5 | 3364 | 16820 |
| 195 | 2 | 7744 | 15488 |
| Total | 30 | | 68280 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{68280}{30} = 2276$$

Answer: Mean $= 107$, Variance $= 2276$

Given: $N = 5+8+15+16+6 = 50$

Step 1: Find midpoints and compute mean

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|---|---|---|---|
| 0–10 | 5 | 5 | 25 |
| 10–20 | 15 | 8 | 120 |
| 20–30 | 25 | 15 | 375 |
| 30–40 | 35 | 16 | 560 |
| 40–50 | 45 | 6 | 270 |
| Total | | 50 | 1350 |

$$\bar{x} = \frac{1350}{50} = 27$$

Step 2: Find $f_i(x_i - \bar{x})^2$

| $x_i$ | $f_i$ | $(x_i - 27)^2$ | $f_i(x_i - 27)^2$ |
|---|---|---|---|
| 5 | 5 | 484 | 2420 |
| 15 | 8 | 144 | 1152 |
| 25 | 15 | 4 | 60 |
| 35 | 16 | 64 | 1024 |
| 45 | 6 | 324 | 1944 |
| Total | 50 | | 6600 |

Step 3: Calculate Variance
$$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{6600}{50} = 132$$

Answer: Mean $= 27$, Variance $= 132$

Given: $N = 3+4+7+7+15+9+6+6+3 = 60$

Let assumed mean $A = 92.5$, $h = 5$, so $y_i = \frac{x_i - 92.5}{5}$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

| Class | $x_i$ | $f_i$ | $y_i$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|
| 70–75 | 72.5 | 3 | $-4$ | $-12$ | 48 |
| 75–80 | 77.5 | 4 | $-3$ | $-12$ | 36 |
| 80–85 | 82.5 | 7 | $-2$ | $-14$ | 28 |
| 85–90 | 87.5 | 7 | $-1$ | $-7$ | 7 |
| 90–95 | 92.5 | 15 | 0 | 0 | 0 |
| 95–100 | 97.5 | 9 | 1 | 9 | 9 |
| 100–105 | 102.5 | 6 | 2 | 12 | 24 |
| 105–110 | 107.5 | 6 | 3 | 18 | 54 |
| 110–115 | 112.5 | 3 | 4 | 12 | 48 |
| Total | | 60 | | 6 | 254 |

Step 2: Find Mean
$$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 92.5 + \frac{6}{60} \times 5 = 92.5 + 0.5 = 93$$

Step 3: Find Variance
$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$
$$= \frac{25}{3600}\left[60 \times 254 - (6)^2\right] = \frac{25}{3600}\left[15240 - 36\right] = \frac{25 \times 15204}{3600} = \frac{380100}{3600} = 105.58$$

Step 4: Standard Deviation
$$\sigma = \sqrt{105.58} \approx 10.28$$

Answer: Mean $= 93$ cm, Variance $\approx 105.58$, Standard Deviation $\approx 10.28$ cm

Given: $N = 15+17+21+22+25 = 100$

Converting to continuous classes: 32.5–36.5, 36.5–40.5, 40.5–44.5, 44.5–48.5, 48.5–52.5

Let assumed mean $A = 44.5$, $h = 4$, so $y_i = \frac{x_i - 44.5}{4}$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

| Class | $x_i$ | $f_i$ | $y_i$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|
| 32.5–36.5 | 34.5 | 15 | $-2.5$ | $-37.5$ | 93.75 |
| 36.5–40.5 | 38.5 | 17 | $-1.5$ | $-25.5$ | 38.25 |
| 40.5–44.5 | 42.5 | 21 | $-0.5$ | $-10.5$ | 5.25 |
| 44.5–48.5 | 46.5 | 22 | $0.5$ | $11$ | 5.5 |
| 48.5–52.5 | 50.5 | 25 | $1.5$ | $37.5$ | 56.25 |
| Total | | 100 | | $-25$ | 199 |

Step 2: Find Mean
$$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 44.5 + \frac{-25}{100} \times 4 = 44.5 - 1 = 43.5 \text{ mm}$$

Step 3: Find Variance and Standard Deviation
$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$
$$= \frac{16}{10000}\left[100 \times 199 - (-25)^2\right] = \frac{16}{10000}\left[19900 - 625\right] = \frac{16 \times 19275}{10000} = \frac{308400}{10000} = 30.84$$

$$\sigma = \sqrt{30.84} \approx 5.55 \text{ mm}$$

Answer: Mean diameter $= 43.5$ mm, Standard Deviation $\approx 5.55$ mm

Given: $n = 8$, $\bar{x} = 9$, $\sigma^2 = 9.25$
Six observations: 6, 7, 10, 12, 12, 13

Let the remaining two observations be $a$ and $b$.

Step 1: Use the mean condition
$$\frac{6+7+10+12+12+13+a+b}{8} = 9$$
$$60 + a + b = 72$$
$$a + b = 12 \quad \cdots (1)$$

Step 2: Use the variance condition
$$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$
$$9.25 = \frac{\sum x_i^2}{8} - 81$$
$$\sum x_i^2 = (9.25 + 81) \times 8 = 90.25 \times 8 = 722$$

Sum of squares of known observations:
$$6^2+7^2+10^2+12^2+12^2+13^2 = 36+49+100+144+144+169 = 642$$

So:
$$642 + a^2 + b^2 = 722$$
$$a^2 + b^2 = 80 \quad \cdots (2)$$

Step 3: Solve equations (1) and (2)
From (1): $(a+b)^2 = 144 \Rightarrow a^2 + 2ab + b^2 = 144$

Using (2): $80 + 2ab = 144 \Rightarrow ab = 32$

So $a$ and $b$ are roots of:
$$t^2 - 12t + 32 = 0$$
$$(t-4)(t-8) = 0$$
$$t = 4 \text{ or } t = 8$$

Answer: The remaining two observations are $4$ and $8$.

Given: $n = 7$, $\bar{x} = 8$, $\sigma^2 = 16$
Five observations: 2, 4, 10, 12, 14

Let the remaining two observations be $a$ and $b$.

Step 1: Use the mean condition
$$\frac{2+4+10+12+14+a+b}{7} = 8$$
$$42 + a + b = 56$$
$$a + b = 14 \quad \cdots (1)$$

Step 2: Use the variance condition
$$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$
$$16 = \frac{\sum x_i^2}{7} - 64$$
$$\sum x_i^2 = 80 \times 7 = 560$$

Sum of squares of known observations:
$$4+16+100+144+196 = 460$$

So:
$$460 + a^2 + b^2 = 560$$
$$a^2 + b^2 = 100 \quad \cdots (2)$$

Step 3: Solve equations (1) and (2)
$(a+b)^2 = 196 \Rightarrow a^2 + 2ab + b^2 = 196$

Using (2): $100 + 2ab = 196 \Rightarrow ab = 48$

So $a$ and $b$ are roots of:
$$t^2 - 14t + 48 = 0$$
$$(t-6)(t-8) = 0$$
$$t = 6 \text{ or } t = 8$$

Answer: The remaining two observations are $6$ and $8$.

Given: $n = 6$, $\bar{x} = 8$, $\sigma = 4$

Each observation is multiplied by 3. Let new observations be $y_i = 3x_i$.

New Mean:
$$\bar{y} = \frac{\sum y_i}{n} = \frac{\sum 3x_i}{n} = 3\bar{x} = 3 \times 8 = 24$$

New Variance:
$$\sigma_y^2 = \frac{1}{n}\sum(y_i - \bar{y})^2 = \frac{1}{n}\sum(3x_i - 3\bar{x})^2 = \frac{9}{n}\sum(x_i - \bar{x})^2 = 9\sigma^2 = 9 \times 16 = 144$$

New Standard Deviation:
$$\sigma_y = \sqrt{144} = 12$$

Answer: New Mean $= 24$, New Standard Deviation $= 12$

Given: $n$ observations $x_1, x_2, \ldots, x_n$ with mean $\bar{x}$ and variance $\sigma^2$.

New observations: $y_i = ax_i$ for $i = 1, 2, \ldots, n$.

Proof of New Mean:
$$\bar{y} = \frac{1}{n}\sum_{i=1}^{n} y_i = \frac{1}{n}\sum_{i=1}^{n} ax_i = a \cdot \frac{1}{n}\sum_{i=1}^{n} x_i = a\bar{x}$$

Hence the mean of the new observations is $a\bar{x}$. $\quad$ Proved.

Proof of New Variance:
$$\sigma_y^2 = \frac{1}{n}\sum_{i=1}^{n}(y_i - \bar{y})^2 = \frac{1}{n}\sum_{i=1}^{n}(ax_i - a\bar{x})^2$$
$$= \frac{1}{n}\sum_{i=1}^{n} a^2(x_i - \bar{x})^2 = a^2 \cdot \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = a^2\sigma^2$$

Hence the variance of the new observations is $a^2\sigma^2$. $\quad$ Proved.

Given: $n = 20$, $\bar{x} = 10$, $\sigma = 2$

So $\sum x_i = 20 \times 10 = 200$ and $\sigma^2 = 4$.

$$\frac{\sum x_i^2}{n} - \bar{x}^2 = 4 \Rightarrow \sum x_i^2 = (4+100) \times 20 = 2080$$

---

(i) Wrong item (8) is omitted:

Correct $\sum x_i = 200 - 8 = 192$, $n = 19$

$$\text{Correct Mean} = \frac{192}{19} \approx 10.1$$

Correct $\sum x_i^2 = 2080 - 64 = 2016$

$$\text{Correct } \sigma^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 = \frac{2016}{19} - \frac{36864}{361}$$
$$= \frac{2016 \times 19 - 36864}{361} = \frac{38304 - 36864}{361} = \frac{1440}{361} \approx 3.99$$

$$\text{Correct } \sigma = \sqrt{\frac{1440}{361}} = \frac{\sqrt{1440}}{19} = \frac{12\sqrt{10}}{19} \approx 1.997 \approx 2$$

---

(ii) Wrong item (8) is replaced by 12:

Correct $\sum x_i = 200 - 8 + 12 = 204$, $n = 20$

$$\text{Correct Mean} = \frac{204}{20} = 10.2$$

Correct $\sum x_i^2 = 2080 - 64 + 144 = 2160$

$$\text{Correct } \sigma^2 = \frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$$

$$\text{Correct } \sigma = \sqrt{3.96} \approx 1.99 \approx 2$$

Answer:
- (i) Correct Mean $\approx 10.1$, Correct S.D. $\approx 2$
- (ii) Correct Mean $= 10.2$, Correct S.D. $\approx 1.99$

Given: $n = 100$, $\bar{x} = 20$, $\sigma = 3$

$\sum x_i = 100 \times 20 = 2000$

$\sigma^2 = 9 \Rightarrow \frac{\sum x_i^2}{100} - 400 = 9 \Rightarrow \sum x_i^2 = 40900$

Step 1: Remove incorrect observations (21, 21, 18)

Correct $n = 100 - 3 = 97$

Correct $\sum x_i = 2000 - 21 - 21 - 18 = 1940$

$$\text{Correct Mean} = \frac{1940}{97} = 20$$

Step 2: Correct $\sum x_i^2$

Correct $\sum x_i^2 = 40900 - 441 - 441 - 324 = 39694$

Step 3: Correct Variance
$$\sigma^2 = \frac{39694}{97} - (20)^2 = 409.216... - 400 = 9.216...$$

More precisely:
$$\sigma^2 = \frac{39694}{97} - 400 = \frac{39694 - 38800}{97} = \frac{894}{97} \approx 9.21$$

$$\sigma = \sqrt{\frac{894}{97}} \approx \sqrt{9.21} \approx 3.036$$

Answer: Correct Mean $= 20$, Correct Standard Deviation $\approx 3.04$