Some psychologists are conducting research with an extraversion inventory, for which, by design, the scores are normally distributed in the general population, with mean 50 and standard deviation 10. What is the probability that the mean extraversion score from a sample of 25 people will be greater than 51?

What is the probability that the mean extraversion score from a sample of 100 people will fall further from the mean (i. e., further out into the tails) than either 49 OR 51 (i. e., add the two areas)?

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

my answer is$ 12

step-by-step explanation:

because 15 -2=12

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)