Question 1
b) Mean = AVERAGE(E2:E252) =1117.64
Standard Deviation = STDEV(E2:E252) = 67.75
c) According to the Empirical Rule with the normal curve being bell-shaped and symmetric about the mean, half of the data is above mean and the other half is below the mean. Therefore the probability is just 0.5.
Question 2
The mean is 1117.64 and the standard deviation is 67.75
The z score value = (950 – 1117.64) / 67.75 = -2.47
The probability P(z>-2.47) = P(z<2.47) = 0.9932
The probability is therefore 0.9932
Question 3
The first z score value = -50/67.75 =- 0.74
The second z score value = 50/67.75 = 0.74
Th probability P(-0.7384.69) = 1 – P(z<4.69) =1 – 0.99999 = 0.00001
The probability is therefore 0.00001
Unusual data is when the values lies more than standard deviations from the mean. Since 800 is less than two standard deviations from the mean it is not considered unusual.
Question 5
(800 – 1117.64) / 67.75 = -4.69
Unusual data is when the values lie more than standard deviations from the mean. Since 800 is less than two standard deviations from the mean it is not considered unusual.
Question 6
Q1 = Q (E2:E252,1) = 1066.2550045 = Q1
Q2 = Q (E2:E252,2) = 1112.200012 = Q2
Q3 = Q (E2:E252,3) = 1173.355011 = Q3
Quartile 7
Properties of normal distribution checks:
Mean = 1117.64 Median =1110.75 Difference = 6.889
Q2 – Q1 = 1112.20 - 1066.26= 45.94
Q3 – Q2 = 1173.36 - 1112.20 = 61.16
The difference between Q1 and Q2 and the difference between Q 2 and Q 3 is not approximately equal. There is a huge difference between the mean and the median.
Therefore, the normality assumption is not valid.