Question

Do these costs appear to come from a population that has a normal distribution? Why or why not? Can the mean of your data sample be treated as a value from a population having a normal distribution? Why or why not? Did an “unusually low” or “unusually high” z-score value occur? Was the associated z score probability value less than 0.05 (p < 0.05); meaning a “significantly low” or “significantly high” event? If yes, what are the implications for the base and/or aircraft? What were your findings? Hint: focus on the calculated mean, standard deviation and z-score (include probability) to interpret your results.

FLYING SQUADRON COSTS Tankers Transpert KC-135 KC-10 C-17 Bomber B-52 Firhters F-16 F-22 LocationLocation Location Altus Wichita Falls Fairehild Castle Jacksonville Barksdale Corpus Christi Clinton Lackland F-15 F-35 Mean Standard Deviation TBD 12.50 10.40 9.30 7.00 4.00 3.10 2.90 2.10 23.40 22.50 11.50 7.10 17.30 7.00 .00 7.00 8.10 6.80 40.70 .10 3.30 Kadena Mildenhall Charleston Miami Enid Hill Beale Mean Standard Deviation 1.10 1.10 0.90 2.60 1.70 44.30 34.20 32.50 38.50 1.30 0.90 0.90 0.60 Score Value Probability

Answer 1

The descriptive statistics of given data are

Variable	Total Count	Mean	StDev	Variance	Skewness
Bomber	6	8.87	4.16	17.29	2.38
F15	5	38.04	4.79	22.98	0.14
Tankers	13	4.43	4.00	15.96	1.06
F35	15	6.02	7.47	55.73	1.80

The F-15 data appears to be normal because the skewness of the data is 0.14. which is slightly left skewed.

Rest others have large skew and they are not approximated to normal distribution.

Considering the total observations as a population, then the population mean = 10.33

z-score = $\frac{\bar{x}-\mu}{\sigma }$

$\mu$ = population mean = 10.33

$\sigma$ = population standard deviation = 12.12

z-score of tankers = $\frac{4.43 - 10.33}{12.12}$ = -0.48 , probability = 0.315

z-score for bombers = $\frac{8.87 - 10.33}{12.12}$ = -0.12 , probability = 0.452

z-score for F-15 = $\frac{38.04 - 10.33}{12.12}$ = 2.286, probability = 0.988

z-score for F-35 = $\frac{6.02 - 10.33}{12.12}$ = -0.355, probability = 0.361

Usually low or high z-score will occur because of large deviation of sample mean from the population mean.