Question

The average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012. The number of times a sample of 20 families dined out last year provides the following data.

Table 4

60	10	50	30	70	3	50	30	10	30
40	10	20	40	10	40	50	60	30	500

1. Compute the mean and median.
2. Compute the first and third quartiles.
3. Compute the range and interquartile range.
4. Compute the variance and standard deviation.
5. The skewness measure for these data is 0.34. Comment on the shape of this distribution. is it the shape you would expect? Why or why not?
6. Do the data contain outliers?

Answer 1

a) Mean is 1143/20= 57.15

The median is middle value when the set is ordered in ascending or descending. The median is an average of two middle values in case of even number of dataset

Median 35

Size n 20

Sum 1143

b) Quartile Q1: 15
Quartile Q3: 50

c)Minimum: 3
Maximum: 500

Range= 500-3 = 497

Interquartile range IQR: 35

Sum of squares deviation=

d) Variance: 10674.3275
Standard deviation σ=103.3166

e) modes are 10,30 here mode is less than mean.

It means that data is right skewed. Hence we expect right skewed distribution of data. So skewness measure 0.34 shows right skewed distribution.

f) yes ,500 is an outlier

Please like ??