# The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceedin

The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.256 ohms and 5% having a resistance smaller than 9.671 ohms. What are the mean value and standard deviation of the resistance distribution? slader

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1. Explanation:

Formula for the given probability is as follows.

$$P(z > \frac{x – \mu}{\sigma})$$ = 0.10

$$P(z > \frac{10.256 – \mu}{\sigma})$$ = 0.10

According to the normal table area we have,

P(z < 1.28) = 0.10

Hence,  $$\frac{10.256 – \mu}{\sigma} = 1.28$$

$$\mu = 10.256 – 1.28 \times \sigma$$ ………. (1)

Also, the given probability is as follows.

$$P(z < \frac{x – \mu}{\sigma})$$ = 0.05

$$P(z < \frac{9.671 – \mu}{\sigma})$$ = 0.05

Hence,     $$\frac{9.671 – \mu}{\sigma} = -1.645$$

$$\mu = 9.671 + 1.645 \times \sigma$$ ……. (2)

Now, substitute the value of $$\mu$$ from equation (1) into equation (2) as follows.

$$10.256 – 1.28 \times \sigma = 9.671 + 1.645 \times \sigma$$

$$-2.925 \sigma = -0.585$$

$$\sigma$$ = 0.2

Putting the value of $$\sigma$$ into equation (2) we will find the value of $$\mu$$ as follows.

$$\mu = 9.671 + 1.645 \times \sigma$$

= $$9.671 + 1.645 \times 0.2$$

= 10

Thus, we can conclude that the value of  $$\sigma$$ is 0.2 and the value of $$\mu$$ is 10.