Question

The Wall Street Journal reported that Walmart Stores Inc. is planning to lay off employees at its Sam's Club warehouse unit. Approximately half of the layoffs will be hourly employees (The Wall Street Journal, January 25-26, 2014). Suppose the following data represent the percentage of hourly employees laid off for Sam's Club stores.55 56 44 43 44 56 60 62 57 45 36 38 50 69 65a. Compute the mean and median percentage of hourly employees being laid off at these stores. Mean Medianb. Compute the first and third quartiles. First quartile Third quartilec. Compute the range and interquartile range. Range Interquartile ranged. Compute the variance and standard deviation. Round your answers to four decimal places. Variance Standard deviatione. Do the data contain any outliers

Answer 1

Explanation:

a) Rearranging the given data in ascending order, it becomes

36, 38, 43, 44, 44, 45, 50, 55, 56, 57, 56, 60, 62, 65, 69

n = 15

The mean = (55 + 56 + 44 + 43 + 44 + 56 + 60 + 62 + 57 + 45 + 36 + 38 + 50 + 69 + 65)/15 = 52

Median = 55

b) Range = 69 - 36 = 33

The median divides the data set into two halves, lower and upper. The median of the lower halve is the first quartile, Q1 while the median of the upper halve is the third quartile, Q3

First quartile, Q1 = 44

Third quartile, Q3 = 60

Inter quartile range, IQR = Q3 - Q1 = 60 - 44 = 16

Variance = (summation(x - mean)²/n

Variance = (55 - 52)^2 + (56 - 52)^2 + (44 - 52)^2 + (43 - 52)^2 + (44 - 52)^2 + (56 - 52)^2 + (60 - 52)^2 + (62 - 52)^2 + (57 - 52)^2 + (45 - 52)^2 + (36 - 52)^2 + (38 - 52)^2 + (50 - 52)^2 + (69 - 52)^2 + (65 - 52)^2/15

Variance = 93.467

Standard deviation = √variance

Standard deviation = √93.467 =

9.668

The data does not contain any outliers.