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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.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 o Assume Intelligent Quotient (IQ) scores are normally distributed with a Mean of 100 and Standard . Deviation of T5 o Assume a friend has an IQ score of 130 (X). o 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 o The probability of another friend scoring higher (non-shaded area) than 130 is: p (1.0000 0.9772) 0.0228 or 2.28%.Day 2 of the airshow arrives, and the weather is worsening. The temperature is 50°F, 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:AIRSHOW -- US MILITARY AIRCRAFT PERFORMANCE Tankers Transport Bomber Fighters Sortie Sortie Airshow Flight Name KC-135 KC-10 C-17 B-52 F-15 F-16 F-22 F-3 Mean Standard Deviation Day Number Sortie 1 Sortie 2 Sortie 3 Sortie 4 Sortie 5 Sortie 6 Sortie 7 Sortie 8 Sortie 9 Sortie 10 Sortie 11 Sortie 12 Sortie 13 Sortie 14 Sortie 15 Sortie 16 Sortie 17 Mean Standard Deviation Z-Score Value Probabilit 2.36 3.70 3.65 3.82 2.30 2.15 3.24 2.50 4.26 2.36 1.80 2.12 2.65 2.00 3.00 2.65 2.45 2.55 3.00 1.75 3.00 1.75 3.00 2.75 2.95 3.10 2.70 1.60 14.71 3.00 3.00 3.81 2.00 4.00 2.65 .00 2.50 3.70 3.00 1.00 1.10 1.05 1.10 1.25 0.98 0.75 0.65 0.45 1.26 0.12 0.13 0.12 0.50 0.98 0.96 0.97 0.92 5.60 1.00 1.00 1.00 1.40 1.15 2 2 2 1.00 1.25 1.15 1.40 3.24 2.50 2 1.05 10.50 1.98 1.96 1.97 1.92

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Answer #1

since the data is not in soft form ( as it is image ) so only using the column KC-135, for the first cell value=2.36 , following calculation is done and question is answered after feeding data into ms-excel.

here n=number of observation and population standard deviation is calculated using all the observation n=16

for for value of 2.36, z=(2.36-3.7831)/2.8718=-0.4956

P(Z<-0.4956)=0.3101 ( you can also use ms-excel command =normsdist(-0.4956))

following information has been generated using

Flight Name KC-135
Stories 1 2.36
Stories 2 2.30
Stories 3
Stories 4 2.45
Stories 5 2.55
Stories 6 3.00
Stories 7 3.00
Stories 8 3.00
Stories 9 2.95
Stories 10 2.70
Stories 11 14.71
Stories 12 3.00
Stories 13 3.00
Stories 14 3.81
Stories 15 4.00
Stories 16 4.00
Stories 17 3.70
n= 16
sum= 60.5300
mean= 3.7831
standard deviation= 2.8718
Z-score value= -0.4956
probability= 0.3101
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