Stat 1040
Chapter 18 Solutions

    1. The average of the box is 4 and the SD is 2.24. The expected value for the sum is 1600 and the standard error is 45. The standard units is -2.22 and the area is about 99%.
    2. The number of "3"s is like the sum of 400 draws from a box with 3 tickets marked "0" and 1 ticket marked "1". The average of the box is .25 and the SD is 0.433. The expected value for the sum is 100 and the standard error is 8.66. The standard units is -1.15 and the area is about 12%.

  2. The box has one ``0'' and one ``1''. The average of the box is 0.5 and the SD is 0.5. The expected value of the sum is 12.5 and the standard error is 2.5. The standard units for the edges of the bar are 1/2.5=0.4 and 0 and the area is approximately 15.54%.

    Note we can solve this even more accurately using the binomial formula of Chapter 15. So the chance is

    [(25!)/(12!13!)]*(1/2)^12*(1/2)^13 = 15.498%.

    (You don't need this many decimal places. I just wanted to show how close the normal approximation is.)

  3. Something is wrong with COIN. The average of the box is 0.5 and the SD is 0.5. The expected value of the sum is 500,000 and the SE is 500. COIN is more than 4 SE's away from the expected value, and that is too many SE's.

  4. (a) is true, because the average of the box is 0.01 and the SD is .0995, so the expected value is 1 and the SE is (approximately) 1. (b) is false because we cannot use the normal curve - the box has too many 0's.