Airshow Day 2: Probability

This discussion introduces you to normal probability via the calculated z-score. A z-score converts a non-standard normal distribution into a standard normal distribution; a standard normal distribution has a mean of zero and standard deviation of one.

Additional z-score properties and details are provided later in the course. For this assignment, what is needed is the capability to calculate a z-score and find its associated probability (see Table A-2 in your textbook). Here is an example:

Excel equation: z = (Your Score (X) – Mean)/(Standard Deviation) OR z = (X – Mean)/S.
Z-score and probability calculation example:
Assume Intelligent Quotient (IQ) scores are normally distributed with a Mean of 100 and Standard Deviation of 15.
Assume a friend has an IQ score of 130 (X).
The z-score is then calculated: z = (130 – 100)/15 = 30/15 = 2.00.
Find Table A-2 in your textbook. The probability (green shaded area) associated with z = 2.00 is p = 0.9772.
The probability of another friend scoring “higher” (non-shaded area) than 130 is: p = (1.0000 – 0.9772) = 0.0228 or 2.28%.
Scenario
Day 2 of the airshow arrives, and the weather is worsening. The temperature is 50oF, and strong thunderstorms are predicted. Continuing intermittent moderate to strong runway crosswinds (25 Knots sustained, with gusts to 40 Knots). Fortunately, all Day 2 flying sorties are accomplished with only minor incidents.

Your team collected these simulated data for Day 2 flying sorties: See attachment

In Microsoft Excel, complete the “Airshow – US Military Aircraft Performance” table by adding the calculated column and row values for mean, standard deviation, and z-score column values (use Sortie 10 data as the value of “X” in the z-score equation); report your calculated values to two decimal places (i.e., 0.12). For reporting probability, use Table A-2 in your textbook and report the value of p to four decimal places (i.e., 0.1234).