The scores of 48 children in an intelligence test are shown in the following frequency table .
`{:("Score ",71,76,79,83,86,89,92,97,101,103,107,110,114),("Frequency ",4,3,4,5,6,5,4,4,3,3,3,2,2):}`
Calculate `sigma^(2)` and find the percentage of students whose score lie between `bar(x) - sigma and bar(x) + sigma `

Answer

To solve the problem, we will follow these steps:

### Step 1: Organize the Data
We have the scores and their corresponding frequencies. Let's summarize them:

| Score (x_i) | Frequency (f_i) |
|-------------|-----------------|
| 71 | 4 |
| 76 | 3 |
| 79 | 4 |
| 83 | 5 |
| 86 | 6 |
| 89 | 5 |
| 92 | 4 |
| 97 | 4 |
| 101 | 3 |
| 103 | 3 |
| 107 | 2 |
| 110 | 2 |
| 114 | 1 |

### Step 2: Calculate the Mean (x̄)
To calculate the mean, we use the formula:
\[ \bar{x} = \frac{\sum (f_i \cdot x_i)}{N} \]
where \(N\) is the total frequency.

Calculating \(N\):
\[ N = 4 + 3 + 4 + 5 + 6 + 5 + 4 + 4 + 3 + 3 + 2 + 2 + 1 = 48 \]

Calculating \(\sum (f_i \cdot x_i)\):
\[ \sum (f_i \cdot x_i) = (71 \cdot 4) + (76 \cdot 3) + (79 \cdot 4) + (83 \cdot 5) + (86 \cdot 6) + (89 \cdot 5) + (92 \cdot 4) + (97 \cdot 4) + (101 \cdot 3) + (103 \cdot 3) + (107 \cdot 2) + (110 \cdot 2) + (114 \cdot 1) \]
Calculating each term:
- \(71 \cdot 4 = 284\)
- \(76 \cdot 3 = 228\)
- \(79 \cdot 4 = 316\)
- \(83 \cdot 5 = 415\)
- \(86 \cdot 6 = 516\)
- \(89 \cdot 5 = 445\)
- \(92 \cdot 4 = 368\)
- \(97 \cdot 4 = 388\)
- \(101 \cdot 3 = 303\)
- \(103 \cdot 3 = 309\)
- \(107 \cdot 2 = 214\)
- \(110 \cdot 2 = 220\)
- \(114 \cdot 1 = 114\)

Now summing these values:
\[ \sum (f_i \cdot x_i) = 284 + 228 + 316 + 415 + 516 + 445 + 368 + 388 + 303 + 309 + 214 + 220 + 114 = 3560 \]

Now, calculating the mean:
\[ \bar{x} = \frac{3560}{48} \approx 74.58 \]

### Step 3: Calculate the Variance (σ²)
Variance is calculated using the formula:
\[ \sigma^2 = \frac{\sum (f_i \cdot d_i^2)}{N} - \bar{x}^2 \]
where \(d_i = x_i - \bar{x}\).

Calculating \(d_i\) and \(d_i^2\):
- For each score, calculate \(d_i\) and \(d_i^2\):

| Score (x_i) | d_i = x_i - 74.58 | d_i^2 | f_i \cdot d_i^2 |
|-------------|---------------------|--------|------------------|
| 71 | -3.58 | 12.8164| 51.2656 |
| 76 | 1.42 | 2.0164 | 6.0488 |
| 79 | 4.42 | 19.5364| 77.4144 |
| 83 | 8.42 | 70.8964| 354.482 |
| 86 | 11.42 | 130.3364| 781.0168 |
| 89 | 14.42 | 207.6964| 1038.482 |
| 92 | 17.42 | 303.6964| 1212.785 |
| 97 | 22.42 | 502.3364| 2009.344 |
| 101 | 26.42 | 698.3364| 2095.012 |
| 103 | 28.42 | 808.6564| 2425.968 |
| 107 | 32.42 | 1055.0564| 2110.112 |
| 110 | 35.42 | 1255.0564| 2510.112 |
| 114 | 39.42 | 1555.0564| 3110.112 |

Now summing \(f_i \cdot d_i^2\):
\[ \sum (f_i \cdot d_i^2) = 51.2656 + 6.0488 + 77.4144 + 354.482 + 781.0168 + 1038.482 + 1212.785 + 2009.344 + 2095.012 + 2425.968 + 2110.112 + 2510.112 + 3110.112 = 18807.04 \]

Now, calculating variance:
\[ \sigma^2 = \frac{18807.04}{48} - (74.58)^2 \]
Calculating \((74.58)^2 \approx 5550.3364\):
\[ \sigma^2 = \frac{18807.04}{48} - 5550.3364 \approx 391.02 - 5550.3364 = 391.02 \]

### Step 4: Calculate Standard Deviation (σ)
\[ \sigma = \sqrt{\sigma^2} \approx \sqrt{391.02} \approx 19.77 \]

### Step 5: Find the Range
Now we find \( \bar{x} - \sigma \) and \( \bar{x} + \sigma \):
\[ \bar{x} - \sigma \approx 74.58 - 19.77 \approx 54.81 \]
\[ \bar{x} + \sigma \approx 74.58 + 19.77 \approx 94.35 \]

### Step 6: Calculate the Percentage of Students
Now, we need to find the number of students whose scores lie between 54.81 and 94.35. From the frequency table, the scores that fall within this range are 71, 76, 79, 83, 86, 89, and 92.

Calculating the total frequency for these scores:
\[ 4 + 3 + 4 + 5 + 6 + 5 + 4 = 31 \]

Now, calculating the percentage:
\[ \text{Percentage} = \left( \frac{31}{48} \right) \times 100 \approx 64.58\% \]

### Final Answers
- Variance (σ²) = 391.02
- Percentage of students whose scores lie between \( \bar{x} - \sigma \) and \( \bar{x} + \sigma \) = 64.58%