2. in a random sample of 130 world campus students, 92 were employed full-time. we want use these data to construct a 95% confidence interval to estimate the proportion of all world campus students who are employed full-time. is it appropriate to use the normal approximation method? show how you checked assumptions.

Step-by-step explanation:

Given that sample size is 130 >30.  Also by central limit theorem, we know that mean (here proportion) of all means of different samples would tend to become normal with mean = average of all means(here proportions)

Hence we can assume normality assumptions here.

Proportion sample given = 92/130 = 0.7077

The mean proportion of different samples for large sample size will follow normal with mean = sample proportion and std error = square root of p(1-p)/n

Hence mean proportion p= 0.7077

q = 1-p =0.2923

Std error = 0.0399

For 95% confidence interval we find that z critical for 95% two tailed is 1,.96

Hence margin of error = + or - 1.96(std error)

= 0.0782

Confidence interval = (p-margin of error, p+margin of error)

= (0.7077-0.0782,0.7077+0.0782)

=(0.6295, 0.7859)

We are 95% confident that average of sample proportions of different samples would lie within these values in the interval for large sample sizes.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl