Answer:
8:6 & 12:9 & 16:12
Step-by-step explanation:
hope this help ☆
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 <em>n</em> = 30.
Thus, the sampling distribution of the sample mean will be normal.
Compute the sample mean and standard deviation as follows:
![\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](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%20x%3D%5Cfrac%7B1%7D%7B30%7D%5Ctimes%2051702%3D1723.4%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x-%5Cbar%20x%29%5E%7B2%7D%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B30-1%7D%5Ctimes%20232561.2%7D%3D89.55)
Construct a 95% confidence interval estimate for the population mean force as follows:
![CI=\bar x\pm z_{\alpha /2}\times\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%20%2F2%7D%5Ctimes%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
![=1723.4\pm 1.96\times\frac{89.55}{\sqrt{30}}\\\\=1723.4\pm 32.045\\\\=(1691.355, 1755.445)\\\\\approx (1691, 1755)](https://tex.z-dn.net/?f=%3D1723.4%5Cpm%201.96%5Ctimes%5Cfrac%7B89.55%7D%7B%5Csqrt%7B30%7D%7D%5C%5C%5C%5C%3D1723.4%5Cpm%2032.045%5C%5C%5C%5C%3D%281691.355%2C%201755.445%29%5C%5C%5C%5C%5Capprox%20%281691%2C%201755%29)
Thus, the 95% confidence interval estimate for the population mean force is (1691, 1755).
Answer:
its answer is c.(16 , 15)
Step-by-step explanation:
June 19th 2006 specifically a monday