Question

The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a resistance smaller than 9.7565 ohms. What are the mean value and standard deviation of the resistance distribution?

Answer 1

Answer:

The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.

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.

10% of all resistors having a resistance exceeding 10.634 ohms

This means that when X = 10.634, Z has a pvalue of 1-0.1 = 0.9. So when X = 10.634, Z = 1.28.

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

$1.28 = \frac{10.634 - \mu}{\sigma}$

$10.634 - \mu = 1.28\sigma$

$\mu = 10.634 - 1.28\sigma$

5% having a resistance smaller than 9.7565 ohms.

This means that when X = 9.7565, Z has a pvalue of 0.05. So when X = 9.7565, Z = -1.96.

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

$-1.96 = \frac{9.7565 - \mu}{\sigma}$

$9.7565 - \mu = -1.96\sigma$

$\mu = 9.7565 + 1.96\sigma$

We also have that:

$\mu = 10.634 - 1.28\sigma$

So

$10.634 - 1.28\sigma = 9.7565 + 1.96\sigma$

$1.96\sigma + 1.28\sigma = 10.645 - 9.7565$

$3.24\sigma = 0.8885$

$\sigma = \frac{0.8885}{3.24}$

$\sigma = 0.2742$

The mean is

$\mu = 10.634 - 1.28\sigma = 10 - 1.28*0.2742 = 9.65$

The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.