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The z-score

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The Standard Normal Distribution

Definition of the Standard Normal Distribution

The Standard Normaldistribution follows a normal distribution and has mean 0 and standard deviation 1

We don't support or recommend that you create a VHDX file that is bootable to both generation 1 and generation 2 virtual machines. The virtual machine generation is a property of the virtual machine, not a property of the virtual hard disk. So you can't tell if a VHDX file was created by a generation 1 or a generation 2 virtual machine. Tap size: NF/NC UNF/UNC: Threads per inch: Basic major dia (inches) Basic effective dia (inches) Basic minor dia of ext. Threads (inches) Basic minor dia of int. Threads (inches). Find P(-1.18 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.

Notice that the distribution is perfectly symmetric about 0.

If a distribution is normal but not standard, we can convert a value to the Standard normal distribution table by first by finding how many standard deviations away the number is from the mean.

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

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

Aiseesoft mac pdf to word converter 3 3 12. 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

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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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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

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.

Example

Find

P(z < 2.37)

2011 macbook pro mojave. 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

Example

Midi keyboard online. 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

Example

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

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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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