In a sample of 80 ten-penny nails, the average weight was

Step1 of 6

Let us consider a random variable \(X\) it represents the weight of the penny nail. Here random variable \(X\) follows normal distribution with mean \(\mu=1.56\) standard deviation \(s = 0.1\)  and \(n = 80\).

That is \(X \sim N\left(\mu, s^{2}\right)\)

\(X \sim N\left(1.56,(0.1)^{2}\right)\)

The probability density function of normal distribution is given by

\(f\left(X \mid \mu, \sigma^{2}\right)=\frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(x-\mu)^{2}}{2 \sigma^{2}}},-\infty<x<\infty\)

Where,

\(x\) = random variable

\(mu\) = mean of X

\(\sigma^{2}\) = variance of \(X\)
\(\sigma\)  = standard deviation of \(X\)

\(\pi\) = mathematical constant and its value is \(3.14\)

Here our goal is:

a). We need to find  95% confidence interval for the mean weight of this type of nail.

b). We need to find 98% confidence interval for the mean weight of this type of nail.

c). We need to check what is the confidence level of the interval (1.54, 1.58)?

d). We need to find how many nails must be sampled so that a 95%confidence interval specifies the mean to within ±0.01 d.

e). We need to find how many nails must be sampled so that a 98% confidence interval will specify the mean to within ±0.01 g.