Answer key for Stat 101 final exam

    1. F(x) = 1 - e-x/10. We want P(X > 15) = 1 - F(15) = 1 - (1 - e-15/10) = e-1.5 = .22.

    2. Note: E(X) = 10, SD(X) = 10

      1. E(Y) = 10,000 x 10 = 100,000 KWH
      2. SD(Y) = sqrt(10,000) x SD(X) = 1000 KWH
      3. P(Y > 102000 KWH) = P((Y-100000)/100 > (102000-100000)/1000) = P(Z > 2) = .025 or more accurately .0228

    3. P(Y > y) = P(Z > (y - 100000)/1000). But we can figure out that P(Z > 2.33) = .01. So, (y - 100000)/1000 = 2.33. Y = 102330 KWH.
    1. (a) increasing the confidence would make the interval wider. Thus to keep it from getting wider, more data must be collected.

    2. (b) The estimate will be biased downward because the previous months are lower than the current month.

    3. (b)

    4. (c) .1 +/- 1.96 sqrt(.1 x .9/100) = .1 +/- .0588

    5. (c) SE = .01. And SE < sqrt(.15 x .85/n). So n = .15 x .85/(.1)2 = 1275. If you use 1.96 instead of 2, you get the exact answer (i.e. start with SE = .02/1.96 instead of SE = .01).

    1. With a sample of 100, we can use either the normal distribution or the t-distribution. The SE = 2/sqrt(100) = .2 minutes. Thus, the confidence interval is 10.5 +/- (1.96) x .2 = [10.1,10.9]. since 10 is not in the confidence interval, it doesn't seem resonable that the mean equals 10.

    2. We want SE = 1/10 minute. But, SE = 2/sqrt(n). So, 1/10 = 2/sqrt(n), so n should be 400. (Again, more accuracy is generate by 1.962 22 / (1/5)2

    3. P(X > 12) = P((X - 10)/2 > (12 - 10)/2) = P(Z > 1). From the empirical rule we guess that this probability should be about 1/6. Using H&O's tables, we get an exact answer of .5 - .3413 = .1587.

      Let N be the number that are greater than 12 out of the 100 sampled items. Then N is a binomial with pi = .1587 and n = 100. Using H&O page 797 we can approximate this by the pi = .15 column which give an answer of .0402 + .0270 + .0171 + .0103 + ... = .09.

      Or we can use the fact that N is approximately a normal with a mean of 15.87 ( = n x pi) and a standard deviation of 3.65 = sqrt(n x pi x (1-pi)). So, P(N >= 20) = P(Z >= 1.15) = .5 - .3749 = .1251. Unfortunately, we also could do the calculation as P(N >= 20) = P(N > 19) = P(Z > .85) = .5 - .3023 = .2977.

    4. Control limits are 9 to 11. (10 +/- 2 x 2/sqrt(16))

    5. About 50%.
      1. Let X be the first estimator. P(X = 11) = P(X = 13) = P(X = 17) = P(X = 19) = 1/4.
      2. by symetry it is unbiased.
      3. SD = 42 x .25 + 22 x .25 + 22 x .25 + 42 x .25 = 10
      1. Let Y be the second estimator. P(Y = 12) = 1/4, P(Y = 15) = 1/2, P(Y = 18) = 1/4.
      2. by symmetry it is unbiased.
      3. SD = 32 x .25 + 02 x .25 + 02 x .25 + 32 x .25 = 4.5

    3. the second estimator has a lower standard deviation.
      1. Y-bar is very skewed. The tail is much longer in the "large values" direction than in the small values direction.

      2. Y-bar: Yes, a CI for its mean owuld be 100 +/- 3.6. So 100 is quite likely the correct mean.

        Complex: Yes. Similar story.

      3. The "complex" estimator has a much smaller standard deviation than Y-bar has. So it is the better estimator.
      1. The average is about 57 degrees. The standard deviation is about .2 degrees.

      2. The temperature has been increasing. It starts out significantly smaller than the average and ends up significantly larger than the average. It doesn't appear that the process is in control.

      Last modified: Wed Dec 22 14:57:39 EST 1999

      -------------------------------------------------------------------------------- 5A) Most got the right. A few said yes, but also emphasised that it was skewed in which case they lost 1 or 3 points depending on their argument. 5B) above 5C) Most argued better sd. If they also argued complex "more normal looking" I took off a point since the test states the distribution is lognormal and so using a normal appearance is an incorrect criterion. 6A) Everyone had the mean correct. I accepted .189 or .194 for the sd. If they got the sd wrong, I subtracted two (+2 for the mean and +1 partial credit for the sd). 6B) I graded this pretty easy. I gave full credit if they had temp increased significantly and/or out of control.