Answer to Question #207956 in Statistics and Probability for jai

Question

 According to a consumer spending survey, the average monthly of BharatBank Visa credit card charge was 1838 INR). A sample of monthly credit card charges provides the following data.

                      236   1710   1351    825   7450

                      316   4135   1333   1584    387

991 3396    170   1428   1688

(ii)  the first and third quartiles.

(iii) the range and interquartile range.

(iv) the variance and standard deviation.

(v)   the skewness measure for these data. Comment on shape of distribution. Is it as expect?

Answer 1

$$170,236,316,387,825,991,1333,1351,$$

$$1428,1584 ,1688,1710,3396,4135,7450$$

(ii)

Since the number of of scores $=15$ is odd, the median is the middle score of the sorted data set: $median =1351.$

The first quartile is the median of the data values below the median (or at 25% of the data)

$$Q_1=387$$

The third quartile is the median of the data values above the median (or at 75% of the data)

$$Q_3=1710$$

(iii)

$$Range=7450-170=7280$$

The interquartile range

$$IQR=Q_3-Q_1=1710-387=1323$$

(iv)

$$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}$$

$$=\dfrac{1}{15}(170+236+316+387+825+991+1333$$

$$+1351+1428+1584+1688+1710+3396$$

$$+4135+7450)=1800$$

$$Var(X)=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{15-1}((170-1800)^2+(236-1800)^2+(316-1800)^2$$

$$(387-1800)^2+...+(7450-1800)^2)$$

$$\approx3675303$$

$$s=\sqrt{s^2}=\sqrt{3675303}\approx1917.1080$$

(v)

$$median<mean$$

The skewness measure is positive. The data are positively skewed or skewed right.

This is to be expected, because most charges will be small, but some will be much larger.