Question

 The data below shows the age distribution of cases of a certain disease reported during the year at a hospital.

34, 17, 25, 37, 19, 19, 27, 19, 44, 24,

24, 22, 32, 12, 13, 16, 18, 14, 12, 16,

14, 17, 10, 16, 22, 20, 15, 15, 10, 10,

14, 17, 20, 18, 19, 13, 13, 13, 18, 30,

24, 34, 44, 31, 43, 40, 28, 31, 18, 22,

15, 31, 18, 27, 35, 35, 20, 32, 38, 32

 Organizing the data into a frequency distribution or table, Calculate the coefficient of Skewness and Kurtosis and interpret your results

Answer 1

The frequency distribution

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} X & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 10 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 12 & 2 & 1/10 & 1/20 & 1/40 \\ \hdashline 13 & 4 & 1/10 & 2/20 & 4/40 \\ \hdashline 14 & 3 & 1/10 & 3/20 & 9/40 \\ \hdashline 15 & 3 & 1/10 & 6/20 & 36/40 \\ \hdashline 16 & 3 & 1/10 & 7/20 & 49/40 \\ \hdashline 17 & 3 & 2/10 & 16/20 & 128/40 \\ \hdashline 18 & 5 & 1/10 & 9/20 & 81/40 \\ \hdashline 19 & 4 & 1/10 & 10/20 & 100/40 \\ \hdashline 20 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 22 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 24 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 25 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 27 & 2 & 1/10 & 14/20 & 196/40 \\ \hdashline 28 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 30 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 31 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 32 & 3 & 1/10 & 14/20 & 196/40 \\ \hdashline 34 & 2 & 1/10 & 14/20 & 196/40 \\ \hdashline 35 & 2 & 1/10 & 14/20 & 196/40 \\ \hdashline 37 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 38 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 40 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 43 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline 44 & 2 & 1/10 & 14/20 & 196/40 \\ \hdashline Total & 60 & 1 & 68/20 & 604/40 \\ \hline \end{array}$$

Mean

$$mean=\bar{X}=\dfrac{1}{60}(10(3)+12(2)+13(4)+14(3)$$

$$+15(3)+16(3)+17(3)+18(5)+19(4)+20(3)$$

$$+22(3)+24(3)+25(1)+27(2)+28(1)+30(1)$$

$$+31(3)+32(3)+34(2)+35(2)+37(1)+38(1)$$

$$+40(1)+43(1)+44(2))=\dfrac{683}{30}=22.76667$$

Variance

$$s^2=\dfrac{1}{60-1}\big(3(10-\dfrac{683}{30})^2+...+2(44-\dfrac{683}{30})^2)$$

$$\approx86.85989$$

Standard deviation

$$s=\sqrt{s^2}=\sqrt{86.85989}\approx9.31987$$

$$median=\dfrac{19+20}{2}=19.5$$

$$Skewness \ Coefficient = \dfrac{3(mean-median)}{s}$$

$$\approx \dfrac{3(\dfrac{683}{30}-19.5)}{9.31987}\approx1.05152$$

The distribution is positively skewed. A positive skew indicates a longer tail to the right.

$$skewness=\dfrac{n\sum_i(X_i-\bar{X})^3}{(n-1)(n-2)s^3}\approx0.69245$$

$$curtosis=\dfrac{n(n+1)\sum_i(X_i-\bar{X})^4}{(n-1)(n-2)(n-3)s^4}$$

$$-3\cdot\dfrac{(n-1)^2}{(n-2)(n-3)}\approx-0.52650$$

A platykurtic distribution shows a negative excess kurtosis. The kurtosis reveals a distribution with flat tails.

Significant skewness and kurtosis clearly indicate that data are not normal.