Question

An electronic parts factory produces resistors. Statistical analysis of the output suggests that resistances follow an approximately Normal distribution with a standard deviation of 0.2 ohms. A sample of 52 resistors has the average resistance of 0.62 ohms. 6.1. [5 pts] Based on these data, construct a 95% confidence interval for the population mean resistance. 6.2. [5 pts] If the actual population mean resistance is exactly 0.6 ohms, what is the probability that an average of 52 resistances is 0.62 ohms or higher

Answer 1

Using the normal distribution and the central limit theorem, it is found that:

a) The 95% confidence interval for the population mean resistance, in ohms, is (0.57, 0.67).

b) 0.2358 = 23.58% probability that an average of 52 resistances is 0.62 ohms or higher.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, the standard deviation for the sampling distribution of sample means of size n is given by $s = \frac{\sigma}{\sqrt{n}}$.

Question a:

We have that:

• Sample of 52 resistors, thus $n = 52$.
• Sample mean of 0.62 ohms, thus $\overline{x} = 0.62$.
• Population standard deviation of 0.2 ohms, thus $\sigma = 0.2$.

The confidence interval is:

$\overline{x} \pm M$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• z is the critical value.
• $\sigma$ is the population standard deviation.
• n is the sample size.

The first step is finding the critical value, which is z with a p-value of $\frac{1 + \alpha}{2}$, in which $\alpha$ is the confidence level.

In this problem, $\alpha = 0.95$, thus, z with a p-value of $\frac{1 + 0.95}{2} = 0.975$, which means that it is z = 1.96.

Then:

$M = 1.96\frac{0.2}{\sqrt{52}} = 0.05$

$\overline{x} - M = 0.62 - 0.05 = 0.57$

$\overline{x} + M = 0.62 + 0.05 = 0.67$

The 95% confidence interval for the population mean resistance, in ohms, is (0.57, 0.67).

Item b:

• Population mean of 0.6 ohms means that $\mu = 0.6$.

• The standard error is:

$s = \frac{0.2}{\sqrt{52}}$.

The probability is 1 subtracted by the p-value of Z when X = 0.62, thus:

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

By the Central Limit Theorem

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

$Z = \frac{0.62 - 0.6}{\frac{0.2}{\sqrt{52}}}$

$Z = 0.72$

$Z = 0.72$ has a p-value of 0.7642.

1 - 0.7642 = 0.2358.

0.2358 = 23.58% probability that an average of 52 resistances is 0.62 ohms or higher.

A similar problem is given at https://brainly.com/question/24663213