4. The values of resistors in a batch follow a normal distribution with a mean of 1 KS2 and a standard deviation of 100 2. Evaluate the probability of a resistor having a resistance a) More than 1.2 k 4. Z=x-M M=1000 5. 5=100. So M DC: 1.2K 0.8K 0.4K 1.IK. 2: +2 -2 -1 +1. a) P(x>1-2K52) = P(=>2) = 1-P(Z<2) = 1 - (0.5 +0.4773) 个 from table -0.0227.
4. The values of resistors in a batch follow a normal distribution with a mean of 1 KS2 and a standard deviation of 100 2. Evaluate the probability of a resistor having a resistance a) More than 1.2 k 4. Z=x-M M=1000 5. 5=100. So M DC: 1.2K 0.8K 0.4K 1.IK. 2: +2 -2 -1 +1. a) P(x>1-2K52) = P(=>2) = 1-P(Z<2) = 1 - (0.5 +0.4773) 个 from table -0.0227.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.3: Special Probability Density Functions
Problem 9E
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this is a practise question given by my lecturer. however they are on holiday so seeking help here. i dont understand why in his given working out that there is a 0.5 for the
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