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:

$\bar x=\frac{1}{n}\sum x=\frac{1}{30}\times 51702=1723.4\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{30-1}\times 232561.2}=89.55$

Construct a 95% confidence interval estimate for the population mean force as follows:

$CI=\bar x\pm z_{\alpha /2}\times\frac{s}{\sqrt{n}}$

$=1723.4\pm 1.96\times\frac{89.55}{\sqrt{30}}\\\\=1723.4\pm 32.045\\\\=(1691.355, 1755.445)\\\\\approx (1691, 1755)$

Thus, the 95% confidence interval estimate for the population mean force is (1691, 1755).

3 0

3 years ago

What is the slope and y-intercept of line p graphed below?

S_A_V [24]

The y int is where the line crosses the y axis ...so ur y int is 3 or (0,3).
ur slope...start at the y int (0,3)....and if u come down 3, and go to the right 2, and down 3, and to the right 2 u are landing on ur line. So ur slope is -3/2

4 0

3 years ago

I need help pls:) and if u need help just tell me

Mariulka [41]

The answer is 28

84/3 =28

8 0

2 years ago