An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $23.1 for a random sample of 1256 people. Assume the population standard deviation is known to be $6. Construct the 90% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

The 90% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $23.4

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 23.1 - 0.3 = $22.8

The upper end of the interval is the mean added to M. So it is 23.1 + 0.3 = $23.4

The 90% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $23.4

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

one that is bigger than 3 1/2