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Though the cumulative standard normal table is most commonly used, your lecturer might instead use the cumulative from zero standard normaltable, which gives the probability that Z is between 0 and a positive number. Graphically this probability is the area under the curve to the left of point "+a" to 0.

The value of "a", called the percentage point, is given along the borders of the table (in bold) and is to 2 decimal places. The values in the main table are probabilities that Z is between 0 and "+a".

Notice values running down the table are to 1 decimal place. The numbers along the column change only for the 2nd decimal place.

Z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
---|---|---|---|---|---|---|---|---|---|---|

0.0 | 0.0000 | 0.0040 | 0.0080 | 0.0120 | 0.0160 | 0.0199 | 0.0239 | 0.0279 | 0.0319 | 0.0359 |

0.1 | 0.0398 | 0.0438 | 0.0478 | 0.0517 | 0.0557 | 0.0596 | 0.0636 | 0.0675 | 0.0714 | 0.0753 |

0.2 | 0.0793 | 0.0832 | 0.0871 | 0.0910 | 0.0948 | 0.0987 | 0.1026 | 0.1064 | 0.1103 | 0.1141 |

0.3 | 0.1179 | 0.1217 | 0.1255 | 0.1293 | 0.1331 | 0.1368 | 0.1406 | 0.1443 | 0.1480 | 0.1517 |

0.4 | 0.1554 | 0.1591 | 0.1628 | 0.1664 | 0.1700 | 0.1736 | 0.1772 | 0.1808 | 0.1844 | 0.1879 |

0.5 | 0.1915 | 0.1950 | 0.1985 | 0.2019 | 0.2054 | 0.2088 | 0.2123 | 0.2157 | 0.2190 | 0.2224 |

0.6 | 0.2257 | 0.2291 | 0.2324 | 0.2357 | 0.2389 | 0.2422 | 0.2454 | 0.2486 | 0.2517 | 0.2549 |

0.7 | 0.2580 | 0.2611 | 0.2642 | 0.2673 | 0.2704 | 0.2734 | 0.2764 | 0.2794 | 0.2823 | 0.2852 |

0.8 | 0.2881 | 0.2910 | 0.2939 | 0.2967 | 0.2995 | 0.3023 | 0.3051 | 0.3078 | 0.3106 | 0.3133 |

0.9 | 0.3159 | 0.3186 | 0.3212 | 0.3238 | 0.3264 | 0.3289 | 0.3315 | 0.3340 | 0.3365 | 0.3389 |

1.0 | 0.3413 | 0.3438 | 0.3461 | 0.3485 | 0.3508 | 0.3531 | 0.3554 | 0.3577 | 0.3599 | 0.3621 |

1.1 | 0.3643 | 0.3665 | 0.3686 | 0.3708 | 0.3729 | 0.3749 | 0.3770 | 0.3790 | 0.3810 | 0.3830 |

1.2 | 0.3849 | 0.3869 | 0.3888 | 0.3907 | 0.3925 | 0.3944 | 0.3962 | 0.3980 | 0.3997 | 0.4015 |

1.3 | 0.4032 | 0.4049 | 0.4066 | 0.4082 | 0.4099 | 0.4115 | 0.4131 | 0.4147 | 0.4162 | 0.4177 |

1.4 | 0.4192 | 0.4207 | 0.4222 | 0.4236 | 0.4251 | 0.4265 | 0.4279 | 0.4292 | 0.4306 | 0.4319 |

1.5 | 0.4332 | 0.4345 | 0.4357 | 0.4370 | 0.4382 | 0.4394 | 0.4406 | 0.4418 | 0.4429 | 0.4441 |

1.6 | 0.4452 | 0.4463 | 0.4474 | 0.4484 | 0.4495 | 0.4505 | 0.4515 | 0.4525 | 0.4535 | 0.4545 |

1.7 | 0.4554 | 0.4564 | 0.4573 | 0.4582 | 0.4591 | 0.4599 | 0.4608 | 0.4616 | 0.4625 | 0.4633 |

1.8 | 0.4641 | 0.4649 | 0.4656 | 0.4664 | 0.4671 | 0.4678 | 0.4686 | 0.4693 | 0.4699 | 0.4706 |

1.9 | 0.4713 | 0.4719 | 0.4726 | 0.4732 | 0.4738 | 0.4744 | 0.4750 | 0.4756 | 0.4761 | 0.4767 |

2.0 | 0.4772 | 0.4778 | 0.4783 | 0.4788 | 0.4793 | 0.4798 | 0.4803 | 0.4808 | 0.4812 | 0.4817 |

2.1 | 0.4821 | 0.4826 | 0.4830 | 0.4834 | 0.4838 | 0.4842 | 0.4846 | 0.4850 | 0.4854 | 0.4857 |

2.2 | 0.4861 | 0.4864 | 0.4868 | 0.4871 | 0.4875 | 0.4878 | 0.4881 | 0.4884 | 0.4887 | 0.4890 |

2.3 | 0.4893 | 0.4896 | 0.4898 | 0.4901 | 0.4904 | 0.4906 | 0.4909 | 0.4911 | 0.4913 | 0.4916 |

2.4 | 0.4918 | 0.4920 | 0.4922 | 0.4925 | 0.4927 | 0.4929 | 0.4931 | 0.4932 | 0.4934 | 0.4936 |

2.5 | 0.4938 | 0.4940 | 0.4941 | 0.4943 | 0.4945 | 0.4946 | 0.4948 | 0.4949 | 0.4951 | 0.4952 |

2.6 | 0.4953 | 0.4955 | 0.4956 | 0.4957 | 0.4959 | 0.4960 | 0.4961 | 0.4962 | 0.4963 | 0.4964 |

2.7 | 0.4965 | 0.4966 | 0.4967 | 0.4968 | 0.4969 | 0.4970 | 0.4971 | 0.4972 | 0.4973 | 0.4974 |

2.8 | 0.4974 | 0.4975 | 0.4976 | 0.4977 | 0.4977 | 0.4978 | 0.4979 | 0.4979 | 0.4980 | 0.4981 |

2.9 | 0.4981 | 0.4982 | 0.4982 | 0.4983 | 0.4984 | 0.4984 | 0.4985 | 0.4985 | 0.4986 | 0.4986 |

3.0 | 0.4987 | 0.4987 | 0.4987 | 0.4988 | 0.4988 | 0.4989 | 0.4989 | 0.4989 | 0.4990 | 0.4990 |

**1. Find p(Z < 0)**

Solution

We want the chance that the variable Z takes a value less than (or less than or equal to) zero.

We do not need the table to find the answer once we know that (1) the area under the curve is 1, and (2) the curve is symmetrical about Z=0, so there's 0.5 (or 50% above Z=0) and 0.5 below Z=0.

**2. Find p(Z < 0.65)**

Solution.

We want the chance that variable Z takes a value less than (or less than or equal to) 0.65.Using common sense we know this number must be bigger than 0.5.

The table gives the probability to the right of Z=0, p(0 < Z < 0.65). To get p(Z < 0.65) just add the area to the left of Z = 0, namely, 0.5 to the value in the table i.e.

p(0 < Z < 0.65) = 0.5 + p( Z < 0.65) = 0.5 + 0.2422 =**0.7422.** กก

Note the value p(Z < 0.65) = 0.2422, is obtained by reading down the rows to 0.6 and then run across the column to 0.05

**3. Find p(Z > 1.12)**

Solution.

Note this time we want the chance Z is GREATER than a number (in this case, 1.12). We can find this by using the fact that P(A) = 1 - P(A). In words, the chance of p(Z > 1.12) is 1 minus the opposite i.e. p(Z < 1.12) i.e.

p(Z > 1.12) = 1 - p(Z < 1.12)

Now before proceeding, get in the habit of thinking "Do I expect the answer to be above 0.5 or below 0.5?" (A sketch of the curve may help.)

Using the table and following the procedure of Example 2, p( Z < 1.12) = 0.5 + p(0 < Z < 1.12) = 0.5+0.3686 = 0.8686

Hence the answer is p(Z > 1.12) = 1 - 0.8686 = **0.1314.**

The answer is below 0.5 which is what we could expect.

**4. Find p(Z < 1.125)**

Solution.

Here you note that the number 1.125 is to 3 decimal places. But the table tabulates to only 2 decimal places? You could use what is called interpolation, but it is far quicker and acceptable to use a bit of common sense. If the value in the 3rd decimal place is

(a) less than five, then round down to 0. Eg 1.32 becomes 1.30 and this makes sense cos the probability for z<1.32 is closer to 1.3 rather than 1.4.

(b) above 5 then round up eg 2.48 becomes 2.50

(c) is 5 then round up or round down or take the average probability of number rounded up and down.

Our case is (c). Let's take the probability for 1.12 and 1.13 and average them for the answer:

p(Z < 1.12) = 0.8686 (from Example 3) and p(Z < 1.13) = 0.5+ 0.3708 =0.8708, so our guess for p(Z < 1.125) is the average of the 2 probabilities: (0.8686+0.8708)/2 = ?

**5. Find p(0 < Z < 1.12)**

Solution.

Here is a case where we want the chance that Z falls within an INTERVAL, and unlike we can read straight off the table, which we would be slightly messier to find if we used the cumulative Z table.

p( 0< Z < 1.12) = **0.3686**

**6. Find p(Z < - 0.52)**

Solution.

Now we want the probability that Z is less than -0.52. We see that the table give values only for positive numbers. (You could have a number with negative numbers but this would be a waster of space as I'll now explain to you.) When you get negative values, you use the idea that the curve is symmetrical about zero.The chance Z < -0.52 is less than 0.5 as can be seen in a sketch.

Using symmetry property, p(Z < -0.52) = 1 - p( Z< 0.52).

And this is p(Z<-0.52) = 1 - [0.5 + p( 0 < Z < 0.52)] = 1 - [ 0.5 + 0.1985 ] = **0.3015**

**7. Find b such that p(Z < b) = 0.975**

Solution.

We want the value "b" such that the probability that Z is less than "b" is 97.5%. What we have here is the reverse problem to those encountered so far. We are given the probability and have to find the percentage point, b.

Since the table gives probability of the interval p( 0 < Z < b) we have to convert the problem to this form. Well p( Z < b) = 0.975 is the same as chopping off the area to the left of Z=0 leaving us with the area above 0 of 0.475 ie

p(Z < b) = 0.975 is the same as p( 0 < Z < b ) = 0.475.

Now look for this probability in the main body of the table, and once you have done this locate the number on the border of the row (1.9) and locate the number in the column border (0.06). So the answer is **b = 1.96**

If you have any difficulties using the table that has not been addressed here, discuss your problem in theforum