Answer to Question #138156 in Statistics and Probability for Ally

Question

Find the measures of variability of the given table. A nutritionist thinks that people with low income average have less Recommended Daily Allowance (RDA) of 800mg intakes of Calcium (standard unit of Calcium for adult). He got a sample of 40 people with income below poverty level. The results are obtained in the following frequency distribution. Intake (mg) Frequency 101-200 1 201-300 1 301-400 9 401-500 13 501-600 10 601-700 6

Answer 1

Solution

The measures of variability are:

Range, Interquartile range, variance, standard deviation and mean deviation.

Range:

Largest value - Smallest value

$$=700-101=599$$

Range is 599

Interquartile Range

$$={Q_3-Q_1}$$

$$Q_1= L+{h \over f}({n\over4}-cf)$$ $$Q_3= L+{h \over f}({3n\over4}-cf)$$

where L= lower limit of the class, h= class width, f= frequency of the class, n= no. of samples, cf= cumulative frequency above the quartile class

${1 \over 4}*40=10 \therefore$ Q₁ lies on the third class between 301-400.

$$\therefore Q_1=300.5+{100 \over 9}({40 \over 4} -2) = 389.38$$

${3 \over 4}*40=30 \therefore Q_3$ lies on the fifth class between 501-600

$$\therefore Q_3=500.5+{100 \over 10}({3*40 \over 4} -24) = 560.5$$

$$IQR={560.5-389.38}=171.12$$

Inter-quartile range is 85.56

Variance

$$var={\sum x^2f -({\sum fx \over n})^2 \over n-1}$$

$$= ({9398810-{18820 \over 40} \over 39})=240983.0641$$

Variance is 240983.0641

Standard deviation

$$SD = \sqrt{var} = \sqrt{240983.0641}$$ $$=490.9002588$$

Standard deviation is 490.9003

Mean deviation

$$={\sum f(x- \bar{x}) \over n} = {0 \over 40}=0$$