Alexis Nguyen

  Question 1: The director of manufacturing at a cookies company needs to determine whether a new machine is able to produce a particular type of cookies according to the manufacturer's specifications, which indicate that cookies should have a mean of 70 and standard deviation of 3.5 pounds. A sample of 49 cookies reveals a sample mean breaking strength of 69.1 pounds. A.  State the null and alternative hypothesis:      Null Hypothesis: The mean strength is 70.     Alternative Hypothesis: The mean strength is not 70.  B.  Is there evidence that the machine is not meeting the manufacturer's specifications for average strength? Use a 0.05 level of significance:      There is no evidence that the machine does not meet the manufacturer's specification for average strength.  We know that the critical values are +1.96 and -1.96. We plug in what we are given to the z formula: Z = (69.1 - 70) / (3.5 / √49) = -0.9 / (3.5 / 7) = -0....

A. Based on Table 1 What is the probability of: B B1 A 10 20 A1 20 40 A1 . Event A =     P(A) = 30/90 = 1/3 A2 . Event B? =   P(B) = 30/90 = 1/3 A3.  Event A or B =  P(A or B) =  P(A) + P(B) - P(A and B)                                                       1/3 + 1/3 - 1/9 = 5/9 A4 . P(A or B) = P(A) + P(B) =   False, because P(A or B) = 5/9 and P(A) + P(B) = 2/3 B1. True B2. We  are given the following:  In terms of probabilities, we know the following: P( A1 ) = 5/365 =0.0136985 [It rains 5 days out of the year.] P( A2 ) = 360/365 = 0.9863014 [It does not rain 360 days out of the year.] P( B | A1 ) = 0.9 [When it rains, the weatherman predicts rain 90% of the time.] P( B | A2 ) = 0.1 [When it does not rain, the weatherman predicts rain 10% of the time.] We wa...