Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{10.634 - \mu}{\sigma}[/tex]
[tex]10.634 - \mu = 1.28\sigma[/tex]
[tex]\mu = 10.634 - 1.28\sigma[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{9.7565 - \mu}{\sigma}[/tex]
[tex]9.7565 - \mu = -1.96\sigma[/tex]
[tex]\mu = 9.7565 + 1.96\sigma[/tex]
We also have that:
[tex]\mu = 10.634 - 1.28\sigma[/tex]
So
[tex]10.634 - 1.28\sigma = 9.7565 + 1.96\sigma[/tex]
[tex]1.96\sigma + 1.28\sigma = 10.645 - 9.7565[/tex]
[tex]3.24\sigma = 0.8885[/tex]
[tex]\sigma = \frac{0.8885}{3.24}[/tex]
[tex]\sigma = 0.2742[/tex]
The mean is
[tex]\mu = 10.634 - 1.28\sigma = 10 - 1.28*0.2742 = 9.65[/tex]
The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.
