Answer:
The 95% confidence interval estimate for the population mean force is (1691, 1755).
Step-by-step explanation:
According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally.
The sample selected here is n = 30.
Thus, the sampling distribution of the sample mean will be normal.
Compute the sample mean and standard deviation as follows:
[tex]\bar x=\frac{1}{n}\sum x=\frac{1}{30}\times 51702=1723.4\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{30-1}\times 232561.2}=89.55[/tex]
Construct a 95% confidence interval estimate for the population mean force as follows:
[tex]CI=\bar x\pm z_{\alpha /2}\times\frac{s}{\sqrt{n}}[/tex]
[tex]=1723.4\pm 1.96\times\frac{89.55}{\sqrt{30}}\\\\=1723.4\pm 32.045\\\\=(1691.355, 1755.445)\\\\\approx (1691, 1755)[/tex]
Thus, the 95% confidence interval estimate for the population mean force is (1691, 1755).