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Examples

Examples. Say a variable has mean 36,500 and standard deviation 5000. What is the probability of getting the value 37,700 or less ? Using the z table we first need to calculate z from the x = 37700.

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Examples

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  1. Examples

  2. Say a variable has mean 36,500 and standard deviation 5000. What is the probability of getting the value 37,700 or less? Using the z table we first need to calculate z from the x = 37700. So z = (37700 – 36500)/5000 = 0.24 which is only 2 decimal places. If it was more you would round the z to 2 decimals. Now in the table a z =.24 has a value .5948 – what does this mean? This means the probability of having a z of .24 or less is 59.48%. So, the table works with the probability of a value or less. What if you wanted to the probability of 37,700 or more? Since the whole distribution adds up to a probability of 1, to get the prob of 37700 or more take 1 minus the prob of 37700 or less. To get this area (probability) take 1 minus tabled value In the table 37700

  3. Now, excel will do most of the work for us if we do not want to use the table. The function =NORMDIST(x value, mean, stand dev, TRUE) gives the area to the left of x. In excel we get the area for the value 37700 as =normdist(37700, 36500, 5000, true) = On the previous slide the table method gave us .5948. Now, we could use excel and just put in the z value using the following =normsdist(z) = normsdist((37700-36500)/5000) = normsdist(.24) = So we have three ways to get the area to the left of a z. I encourage you to use the table first and foremost because that will be easiest to use on tests.

  4. Say you want the area between a and b. You would find the area totally to the left of b and subtract the area totally to the left of a. The area to the left of b is q+p and the area to the left of a is q, so the area between a and b is q+p-q=p. q p a b

  5. Problem 6.2 page 192 .9671 - .0582 = .9089 .0582 1-.9671 =.0329 Z -1.57 1.84 a. We see Z’s -1.57 and 1.84. The area to the left of -1.57 is .0582 and the area to the left of 1.84 is .9671. We want what is between -1.57 and 1.84. You see that is .9089. b. To the left of -1.57 or greater than 1.84 is found by adding those two areas .0582 + .0329 = .0911

  6. Problem 6.2 page 192 1 - .025 = .9750 .025 Z c. Here we have to find a z value such that the “upper tail” area is .025. The table gives areas to the left of a value, and we know our area to the left should be 1 - .025 = .9750 We see the Z = 1.96 has this value exactly! So only .025 Z values are larger than 1.96

  7. Problem 6.2 page 192 .6826/2 = .3413 .6826/2 = .3413 .5 - .3413 = .1587 Z Zl Zr 0 d. The Z table, the table of the standard normal distribution, is symmetric. This means we take the .6826 and divide it an half and we want to focus on the Z’s that give us the area .3413 on each side of 0. Note that half the area is to the left of 0. On the left side we want Zl. From 0 all the way back on the left we have area = .5. The table deals with areas to the left of a value. The area to the left of Zl has to equal .5 minus .3413 = .1587 for us here. It turns out in the table the Z = -1.00 has area .1587. Zr should have area = .1587 + .3413 + .3413 = .8413. This Z = 1.00.

  8. Problem 6.6 page 193 I did’t draw in the normal curve here. Will you draw it in, please? Make the highest point above 50 43 50 Z a. P(X > 43) since the normal distribution is continuous when we look at P(X > 43) we just start at 43 and look to the right. If it was discrete we would go to 44, but since it is continuous we would miss all the values from 43 to 44. We have to translate the X of 43 to a Z. Z = (43-50)/4 = -7/4 = -1.75. In the table with a Z = -1.75 we get the area to the left as .0401. But, we want the area to the right. So P(X > 43) = 1 - .0401 = .9599

  9. Problem 6.6 page 193 b. P(X < 42) Since we just want the area to the left of 42 we translate to Z = (42 – 50)/4 = -8/4 = -2.00. The area to the left is .0228. So, P(X < 42) = .0228. c. What X when translated to a Z has area to the left of .0500? Note a Z = -1.64 has area .0505 and a Z = -1.65 has area .0495. .0500 is exactly in the middle of .0505 and .0495. In this case there is a tradition to use a Z = -1.645. Now from the Z formula -1.645 = (X – 50)/4, or -1.645(4) = X – 50, or 50 – 1.645(4) = X, or X = 50 – 1.645(4) = 43.42

  10. Problem 6.6 page 193 You draw in the normal distribution, please. .2 .3 .3 .2 Xl 50 Xr d. Xl will correspond to a Z that has .2000 in the table. The closest Z = -0.84 (with an area to the left of .2005). The X then is found from -.84 = (X – 50)/4, or X = 50 - .84(4) = 46.64. Xr will correspond to a Z that has .8000 in the table. The closest is .84 (with an area to the left of .7995). The X is found from .84 = (X – 50)/4, or X = 50 + .84(4) = 53.36

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