You collect the following data on the air pressure of footballs ready for play. No data points are outside of the specification limits, so it appears the process is under control. However, you want to calculate the actual probability of a defect occurring in the future. Calculate this probability assuming the standard deviation does not change and that Tom Brady will not deflate the footballs right before the game.

0.00086 or 0.086%

Step-by-step explanation:

The question is not complete, however, after searching online for the question, I was able to get the complete question.

Given that

USL = 14 ; CL = 13 ; LSL = 12

Probability of air pressure > 14 is equal to the  Probability of air pressure <12 because the process is centered

Therefore, the probability of defect in future is equal to 2 × Probability of air pressure > 14

For us to discover the probability of pressure > 14, finding Z-value for X= 14 is important,

Mean Pressure (mu) =13

Std. Deviation (Sigma) = 0.3

Therefore;

Z = (X-mu)/sigma = (14-13)/0.3 = 3.33

From Standard normal distribution table for Z = 3.3, p = 0.99957

Hence,

the probability of pressure > 14 = 1 - 0.99957 = 0.00043

Finally, the probability of defect in future

= 2 × Probability of air pressure > 14

= 2 × 0.00043

= 0.00086

= 0.086 %

answer: 59

step-by-step explanation:

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