Mean	1930.116
Standard Error	558.8155
Median	945
Mode	340
Standard Deviation	3831.046
Sample Variance	14676912
Kurtosis	32.42707
Skewness	5.366171
Range	25365
Minimum	135
Maximum	25500
Sum	90715.45
Count	47
Confidence Level(90.0%)	938.0614

a.

(1) Mean greater than the median, which indicates that data is right skewed and not normally distributed.

(2) Skewness is 5.37. A positive skewness of 5.37 suggests that mean is not equal to median. Hence, the data is not normally distributed.

(3) Difference between Median and minimum reading (945 - 135 = 810) is very less as compared to difference between maximum reading and the median( 25500 - 945 = 24555). In normal distribution, the difference to be roughly same. Hence, the data is right skewed.

(4) Kurtosis, (32.42707 0).

(5) Range 6s (25365 6 x 3831.046).

Therefore, these are five places of evidence which shows that the data is not normally distributed.

b.

PZ< 1.5

= P - 1.5 < Z < 1.5

= PZ < 1.5 - PZ < -1.5

= 1-PZ<-1.5-PZ<-1.5 = 1- 2P(Z<-1.5) = PZ<1.5 = 1- 20.0668=0.8664 = PZ<1.5=0.87=87% In my sample, 87% of observations are expected to be within 1.5 standard deviations of the mean.

(c) Use the mean and standard deviation from the Descriptive Statistics table of Task 4 to calculate the bound for 1.5 standard deviation spread from the mean. Using the "Sold Price" sample data, manually count the number of observations which falls within the bound. State whether this count matches, approximately, your answer to (b) and hence whether this result confirms (or not) your conclusion in (a).