Assume the sample variances to be continuous measurements. Find the probability that a random sample of 36 observations, from a normal population with variance sigma^2 = 6, will have a sample variance S^2 a) greater than 9.1;
b) between 3.5 and 10.5.

0.02566

0.972559

Step-by-step explanation:

Given that :

Sample size (n) = 36

σ² = 6

sample variance S^2

a) greater than 9.1;

b) between 3.5 and 10.5.

The degree do freedom (df) = n - 1

S² > 9.1

P(S² > 9.1) = X² > ((n - 1) * S²) / 6

P(S² > 9.1) = ((36 - 1) * 9.1) / 6

P(S² > 9.1) = (35 * 9.1) / 6

P(S² > 9.1) = 53.083

P(X² > 53.08) = 0.02566 ( chi square distribution calculator)

b) between 3.5 and 10.5.

((36 - 1) * 3.5) / 6 ≤ S² ≤ ((36 - 1) * 10.5) / 6

20.416666 ≤ S² ≤ 61.25

Using the Chisquare distribution calculator :

P(X² > 20.42) = 0.9765

P(X² > 61.25) = 0.003941

Hence,

0.9765 - 0.003941 = 0.972559

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!