Xx=3 f (x) = 2 c + c + 2 c = 5 c c = P ({x = 4} {x 4}) P (x 4) f (4) f (4) + f (5) P (x ) = p (x 4) = f (4) + f (5) = 2. (a) the histogram is approximately symmetric and there is a linear ten- dency in the normal probability plot. It is reasonable to assume that the tar content is normally distributed. (b) we want to test h0 : = 14 against h1 : > 14. The observed value of the test statistic is t0 = x 14 s/ n. The p-value : p = p (t > 2. 894), where t has a t(24) distribution. From the table for the t distribution, we obtain 0. 0025 < p-value < We have su cient evidence to conclude that > 14 at = 5%. (c) a 95% con dence interval for the is x t0. 025,24 s.