The scores on a placement test given to college freshmen for the past five years are approximately normally distributed with a mean μ = 74 and a variance σ2 = 8. would you still consider σ2 = 8 to be a valid value of the variance if a random sample of 20 students who take the placement test this year obtain a value of s2 = 20?

And on this case we have a 0.0303% of chance that the sample variance would be higher than 20. So on this case its not reasonable that the sample variance would be higher than 20.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

represent the sample size

represent the sample variance obtained

represent the population variance

We can calculate the probability that would be higher than 20 and we will have an idea how to solve the problem.  

For this case we can use the following statistic:

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

Calculate the probability

We can calculate the probability that would be higher than 20. The degrees of freedom are given by n-1=20-1=19

And hte reason of this is this one:

In order to calculate this probability we can use the following code in excel:

"==1-CHISQ.DIST(47.5,19,TRUE)"

And on this case we have a 0.0303% of chance that the sample variance would be higher than 20. So on this case its not reasonable that the sample variance would be higher than 20.

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3