Question

5) Two machines M1, M2 are used to manufacture resistors with a design

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 1050, \sigma = 100$

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1100 - 1050}{100}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915.

X = 900

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{900 - 1050}{100}$

$Z = -1.5$

$Z = -1.5$ 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 $\mu = 1000, \sigma = 120$

X = 1100

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1100 - 1000}{120}$

$Z = 0.83$

$Z = 0.83$ has a pvalue of 0.7967.

X = 900

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{900 - 1000}{120}$

$Z = -0.83$

$Z = -0.83$ 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.