Chocolate chip cookies have a distribution that is approximately normal with a m

Question

Chocolate chip cookies have a distribution that is approximately normal with a mean of 23.7 chocolate chips per cookie and a standard deviation of 2.8 chocolate chips per cookie. For quality control purposes, a manufacturer wants to make sure that the number of chocolate chips stays within reasonable limits. Production has been set to keep the number of chips at or below the 90th percentile. Therefore, any cookie with a chip count in the top ten percentile will be set aside for evaluation. What is the maximum number of chips that a cookie should have to meet the requirements of quality control? Use the information above to find the Z SCORE that you will use in this problem: Use the Z SCORE to calculate the chip count that separates the bottom 90. from the top 10%

Explanation / Answer

Normal Distribution
Mean ( u ) =23.7
Standard Deviation ( sd )=2.8
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 23.7/2.8 ) = 0.9
That is, ( x - 23.7/2.8 ) = 1.28
--> x = 1.28 * 2.8 + 23.7 = 27.2896                  

the chip count that separates the bottom 90. from the top 10% is 27.29

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