The answer to this question has to be 8
Since in this case we are
only using the variance of the sample and not the variance of the real population,
therefore we use the t statistic. The formula for the confidence interval is:
<span>CI = X ± t * s / sqrt(n) ---> 1</span>
Where,
X = the sample mean = 84
t = the t score which is
obtained in the standard distribution tables at 95% confidence level
s = sample variance = 12.25
n = number of samples = 49
From the table at 95%
confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t
is around:
t = 1.68
Therefore substituting the
given values to equation 1:
CI = 84 ± 1.68 * 12.25 /
sqrt(49)
CI = 84 ± 2.94
CI = 81.06, 86.94
<span>Therefore at 95% confidence
level, the scores is from 81 to 87.</span>
B B B
B is the correct answer
I’m not exactly sure but maybe f(0) = 4 and g(–2) = 4
