Chapter 4, Problem 4.1P

To determine

Range of values containing 70% of the population of x

Answer to Problem 4.1P

Range of values $5.066\le x\le 7.134$

Explanation of Solution

Given:

  $AssumeNumberofsample$

  $N=1000$

  $SampleMean$

  $\overline{x}=6.1$

  $units$

  $SampleMeanStandarddeviation$

  $\sigma =1$

Concept Used:

  $TestHypothesesbyztest:z_0=\frac{\overline{x}-\mu }{\sigma }$

Interval defined: $\stackrel{-}{x}-z_0\sigma \le x_j\le \stackrel{-}{x}+z_0\sigma$

Calculation:

As per the given problem

Assuming that the sample data is adequately large such that its population behaves as an infinite population, so it needs to solve the integral

  $p\left(z_0\right)=\frac{1}{\sqrt{2\pi }}\underset{0}{\overset{z_1}{{\displaystyle \int }}}{e}^{-{\beta }^2/2}d\beta$ ............(1)

Over the interval defined by $z_0$ . Table 4.3 lists the solutions for this integral. Using Table 4.3 for, we find $p\left(z_0\right)=0.35 ,$ Then the value of $z_0$ =1.034.We should expect that 70% of the xjvalues lie in the interval given by

  $\stackrel{-}{x}-z_0\sigma \le x_j\le \stackrel{-}{x}+z_0\sigma$

  $6.1-1.034\le x_j\le 6.1+1.034$

  $5.066\le x_j\le 7.134$