Answer:
Since M1 has the higher probability of being in the desired range, we choose M1.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.
So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.
For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.
M1:
Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that ![\mu = 1050, \sigma = 100](https://tex.z-dn.net/?f=%5Cmu%20%3D%201050%2C%20%5Csigma%20%3D%20100)
The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So
X = 1100
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{1100 - 1050}{100}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1100%20-%201050%7D%7B100%7D)
![Z = 0.5](https://tex.z-dn.net/?f=Z%20%3D%200.5)
has a pvalue of 0.6915.
X = 900
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{900 - 1050}{100}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B900%20-%201050%7D%7B100%7D)
![Z = -1.5](https://tex.z-dn.net/?f=Z%20%3D%20-1.5)
has a pvalue of 0.0668
0.6915 - 0.0668 = 0.6247.
M1 has a 62.47% probability of being in the desired range.
M2:
M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that ![\mu = 1000, \sigma = 120](https://tex.z-dn.net/?f=%5Cmu%20%3D%201000%2C%20%5Csigma%20%3D%20120)
X = 1100
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{1100 - 1000}{120}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1100%20-%201000%7D%7B120%7D)
![Z = 0.83](https://tex.z-dn.net/?f=Z%20%3D%200.83)
has a pvalue of 0.7967.
X = 900
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{900 - 1000}{120}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B900%20-%201000%7D%7B120%7D)
![Z = -0.83](https://tex.z-dn.net/?f=Z%20%3D%20-0.83)
has a pvalue of 0.2033
0.7967 - 0.2033 = 0.5934
M2 has a 59.34% probability of being in the desired range.
Since M1 has the higher probability of being in the desired range, we choose M1.