A sample of 30 randomly selected oranges was taken from a large population, and their diameters were measured. The mean diameter of the sample was 91 mm and the standard deviation was 8 mm. Assuming a Normal distribution, calculate (correct to one decimal place) 85% confidence limits for the mean diameter of the whole population of oranges.

A 87.9 mm and 94.1 mm

B 89.9 mm and 93.1 mm

C 88.1 mm and 93.9 mm

D 88.9 mm and 93.1 mm

D 88.9 mm and 93.1 mm

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 sample mean subtracted by M. So it is 91 - 2.1 = 88.9mm

The upper end of the interval is the sample mean added to M. So it is 91 + 2.1 = 93.1 mm

So the correct answer is:

D 88.9 mm and 93.1 mm

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.