1. Mean Calculation

To estimate the mean, we’ll use the midpoint for each interval, multiply it by the frequency of that interval, and then sum all the values and divide by the total number of countries.

Step 1: Find Midpoints

• For each interval, find the midpoint (average of lower and upper bound).

Interval	Midpoint (x)	Frequency (f)
16.1 – 21.5	18.8	21
21.6 – 25.9	23.75	7
26.0 – 30.3	28.15	7
30.4 – 35.7	33.05	3
35.8 – 37.3	36.55	3
37.4 – 42.5	39.95	3
42.6 – 48.7	45.65	2

Step 2: Multiply Midpoints by Frequencies

• Calculate $f\times xf\; \backslash times\; x$for each interval.

Interval	Midpoint (x)	Frequency (f)	$f\times xf\; \backslash times\; x$
16.1 – 21.5	18.8	21	394.8
21.6 – 25.9	23.75	7	166.25
26.0 – 30.3	28.15	7	197.05
30.4 – 35.7	33.05	3	99.15
35.8 – 37.3	36.55	3	109.65
37.4 – 42.5	39.95	3	119.85
42.6 – 48.7	45.65	2	91.3

Step 3: Calculate Mean

• Sum the products $f\times xf\; \backslash times\; x$f×x and divide by the total frequency.

$Mean=\frac{\sum (f\times x)}{\sum f}\backslash text\{Mean\}\; =\; \backslash frac\{\backslash sum\; (f\; \backslash times\; x)\}\{\backslash sum\; f\}$​ $Mean=\frac{394.8+166.25+197.05+99.15+109.65+119.85+91.3}{21+7+7+3+3+3+2}\backslash text\{Mean\}\; =\; \backslash frac\{394.8\; +\; 166.25\; +\; 197.05\; +\; 99.15\; +\; 109.65\; +\; 119.85\; +\; 91.3\}\{21\; +\; 7\; +\; 7\; +\; 3\; +\; 3\; +\; 3\; +\; 2\}$​ $Mean=\frac{1177.05}{46}\approx 25.59\backslash text\{Mean\}\; =\; \backslash frac\{1177.05\}\{46\}\; \backslash approx\; 25.59$

2. Standard Deviation Calculation

Step 1: Calculate $(x-mean{)}^2(x\; –\; \backslash text\{mean\})^2$(x−mean)2 for Each Interval

• Find the deviation of each midpoint from the mean and square it.

Step 2: Multiply by Frequency

• Multiply each squared deviation by the frequency for each interval.

Interval	Midpoint (x)	$(x-25.59{)}^2(x\; –\; 25.59)^2$	Frequency (f)	$f\times (x-25.59{)}^{2}f\; \backslash times\; (x\; –\; 25.59)^2$
16.1 – 21.5	18.8	46.13	21	968.73
21.6 – 25.9	23.75	3.38	7	23.66
26.0 – 30.3	28.15	6.59	7	46.13
30.4 – 35.7	33.05	55.47	3	166.41
35.8 – 37.3	36.55	119.71	3	359.13
37.4 – 42.5	39.95	206.96	3	620.88
42.6 – 48.7	45.65	403.53	2	807.06

Step 3: Sum and Divide by Total Frequency

• Sum the products and divide by the total frequency, then take the square root.

$Variance=\frac{\sum (f\times (x-mean{)}^2)}{\sum f}\backslash text\{Variance\}\; =\; \backslash frac\{\backslash sum\; (f\; \backslash times\; (x\; –\; \backslash text\{mean\})^2)\}\{\backslash sum\; f\}$​ $Variance=\frac{2992.00}{46}\approx 65.04\backslash text\{Variance\}\; =\; \backslash frac\{2992.00\}\{46\}\; \backslash approx\; 65.04$ $Standard Deviation=\sqrt{65.04}\approx 8.06$

3. Determine Unusual Intervals

We consider values more than two standard deviations away from the mean to determine if an interval is unusual.

• Mean = 25.59
• Standard Deviation = 8.06
• $2\times 8.06=16.122\; \backslash times\; 8.06\; =\; 16.12$
  

Unusual Range:

• Below: $25.59-16.12=9.4725.59\; –\; 16.12\; =\; 9.47$
• Above: $25.59+16.12=41.7125.59\; +\; 16.12\; =\; 41.71$

Unusual Intervals:

• The interval 42.6 – 48.7 is above 41.71, making it unusually high.

Conclusion

• Mean Percentage of Underweight Children: 25.59
• Standard Deviation: 8.06
• Unusual Intervals:
  • The intervals 37.4-42.5 and 42.6-48.7 could be considered unusually high since they contain values that are at least two standard deviations above the mean percentage of underweight children.