The z-score

The number of standard deviations from the mean is called the z-score and can be found by the formula

x – m

z = ——–

s

x is the value

m is "mu", the mean

s is standard deviation

Example

Find the z-score corresponding to a raw score of 132 from a normal distribution with mean 100 and standard deviation 15.

Solution

We compute

132 – 100

z = ————- = 2.133

15

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Example

A z-score of 1.7 was found from an observation coming from a normal distribution with mean 14 and standard deviation 3. Find the raw score.

Solution

We have

x – 14

1.7 = ———-

3

To solve this we just multiply both sides by the denominator 3,

(1.7)(3) = x – 14

5.1 = x – 14

x = 19.1

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The z-score and Area

Often we want to find the probability that a z-score will be less than a given value, greater than a given value, or in between two values. To accomplish this, we use the table from the textbook and a few properties about the normal distribution. There are three cases; less than, greater than and between.

**Example for less than**

Find

P(z < 2.37)

Solution

We use the table. Notice the picture on the table has shaded region corresponding to the area to the left (below) a z-score. This is exactly what we want. Below are a few lines of the table.

z |
.00 |
.01 |
.02 |
.03 |
.04 |
.05 |
.06 |
.07 |
.08 |
.09 |

2.2 |
.9861 | .9864 | .9868 | .9871 | .9875 | .9878 | .9881 | .9884 | .9887 | .9890 |

2.3 |
.9893 | .9896 | .9898 | .9901 | .9904 | .9906 | .9909 | .9911 |
.9913 | .9916 |

2.4 |
.9918 | .9920 | .9922 | .9925 | .9927 | .9929 | .9931 | .9932 | .9934 | .9936 |

The columns corresponds to the ones and tenths digits of the z-score and the rows correspond to the hundredths digits. For our problem we want the row 2.3 (from 2.37) and the row .07 (from 2.37). The number in the table that matches this is .9911.

Hence

P(z < 2.37) = .9911

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**Example for greater than**

Find

P(z > 1.82)

Solution

In this case, we want the area to the right of 1.82. This is not what is given in the table. We can use the identity

P(z > 1.82) = 1 – P(z < 1.82)

reading the table gives

P(z < 1.82) = .9656

Our answer is

P(z > 1.82) = 1 – .9656 = .0344

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**Example for in between**

Find

P(-1.18 < z < 2.1)

Solution

Once again, the table does not exactly handle this type of area. However, the area between -1.18 and 2.1 is equal to the area to the left of 2.1 minus the area to the left of -1.18. That is

P(-1.18 < z < 2.1) = P(z < 2.1) – P(z < -1.18)

To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table to get

P(z < 2.10) = .9821.

The table also tells us that

P(z < -1.18) = .1190

Now subtract to get

P(-1.18 < z < 2.1) = .9821 – .1190 = .8631

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