# standard normal distribution curve

As you know the normal distribution curve is defined by mean µ and standard deviation σ. The mean can take any value and the standard deviation can take any positive value. The Standard Normal Distribution curve is a normal distribution with a mean of zero and a standard deviation of one.

A standard normal distribution curve is used to determine the probability that a variable following a normal distribution will take values in a specified range. Suppose we study a variable such as daily production errors, the heights of a group of people, or the time of a process, and we find that it follows a normal distribution with a mean of 35 and a standard deviation of 2, and we want to estimate the probability that the value of this variable is greater than 40. We need tables that show the area under this The curve because this area – as we showed in the previous article – expresses the possibilities. So we would need a table for each normal distribution curve, which is very complicated. So we use a simple equation to convert the variable value of the standard distribution curve and thus we can use only one table which is the standard normal distribution curve.

The conversion process from any normal distribution to the standard normal distribution is done using a simple equation, where we denote the original variable with X and its counterpart in the standard (standard) curve with Z. The conversion is done using the following equation:

where μ is the mean and σ is the standard deviation. In the previous example, the value of Z corresponding to X = 40 is

(40 – 35) 2 = 2.5

Thus, we search in the standard normal distribution table for a value of 2.5, which we find corresponds to 0.993, meaning that the area on the left is equal to this value, which corresponds to X being less than 40. But we are looking for a probability of X greater than 40. Therefore, we are looking for the area on the right of the curve, which is 1- 0.993 = 0.017. That is, the probability that X will exceed forty is 1.7%.

Note that the total area under the normal distribution curve is equal to 1 in all cases, so we subtracted the value we got from 1 to get the area to the right of the curve.

For the same result using Excel or Calc using the NORMSDIST function, so we write in any cell

NORMSDIST(2.5) = 0.993

But we have to note that this is the area to the left of 2.5, as it means the probability that X is less than 40.

Is it possible to determine the probability that X is between 30.5 and 32? Yes, we have to calculate the area under the curve to the left of each value and then subtract them to get the area between these two values which, as you know, is equal to the probability of X falling between these two values.

Z1= (30.5 – 35) 2 = -2.25

Z2 = (32 – 35) 2 = – 1.5

Using tables or computers, we find that the areas are 0.122 and 0.066, and the difference between them is 0.054, meaning that the probability of X falling between 30.5 and 32 is 5.4%.

Transformation Concept of Standard Normal Distribution Curve:

The idea of using a standard normal distribution curve to calculate the probabilities of non-standard normal curves may seem strange and inconspicuous, but it’s actually like a lot of things you’ve come across before. The conversion process for a standard normal distribution curve is similar to measuring an area in square inches and then using a conversion factor to convert it to square metres. It is also similar to drawing very large countries on a small map using scale, then measuring distances from the map and converting them to their original value using scale. It can be likened to measuring the area of the figure below using the area of small rectangles whose area is 1 square centimeter, so we find that the area is equal to 14 square centimetres.

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The standard distribution curve is a way to calculate the probabilities (area under the curve) of any normal distribution curve. We can convert the value (X) of any variable that follows a non-standard normal distribution to its counterpart (Z) in the standard normal distribution curve and thus we can estimate the area under the curve. Converting from X to Z and back is similar to using scale on maps. Calculating the area under the first curve using the area under the standard curve is similar to measuring the area of a shape using small squares of known area.

The figure below shows an example of the conversion process. We have a normal distribution with a mean = 15 and a standard deviation of 3. We want to estimate the probability that this variable falls between 16 and 20. We use the transformation to convert the values of 16 and 20 to their standard distribution counterparts of 0 and 1.33. What does this conversion mean? The meaning of this conversion is that the area that we originally want to calculate, colored in green and located below the original curve between values 16 and 20, is equal to the area under the standard curve between values 0 and 1.33, colored in red, despite the difference in shape. Thus, the conversion enables us to estimate the area colored in red using standard normal distribution tables or using a computer. Thus, we have reached the original area (green), which expresses the possibility that the value of the variable under study is between 16 and 20. In this example, we find that this area is equal to 0.40, meaning that the area between 0 and 1.33 in the standard curve is equal to 0.40, which is equal to the area under the curve The original is between 16 and 20, which means that the probability of the variant falling between 16 and 20 is 40%.

Examples:

Example 1: Suppose the preparation time for a drink in a restaurant varies from time to time with a mean of 2 minutes and a standard deviation of 0.5 minutes. What is the probability that the brew time is less than 3 minutes?

First we calculate the value of Z which is equivalent to X

Z = (3-2) / 0.5 = 2

Using tables or computers, we find that the area under the curve to the left of the value 3 (red) is equal to 97.7%, meaning that the probability that the preparation time of the drink is less than 3 minutes is 97.7%.

We can conclude that the probability that the drink preparation time is greater than 3 minutes is 1 – 97.7% = 2.3%.

Example 2: Suppose the length of an item being produced is 60 cm and the customer orders the length to be in the range of 59.95 cm and 60.08 cm. By following up the production process, we found that we produce the piece with an average of 59.99 cm and a standard deviation of 0.04 cm. What is the probability of exceeding the tolerance allowed by the customer

The figure below shows the normal distribution curve, which represents the change in the length of this piece in production. What is required is to calculate the area to the right of 60.08 (green) and the area to the left of 59.95 (red).

We calculate the Z value which is equivalent to 59.95 and we find it

Z = (59.95 – 59.99) / 0.04 = -1

Using tables or computers, we find that the area to the left of this value is 15.87%. Is this the value we are looking for or should we subtract it from 1 like we did in the previous example? We are looking for the probability that the length is less than this value, because we really want the area to the left of this value.

Then we calculate the Z value which is equivalent to 60.08 and find it

Z = (60.08- 59.99) / 0.04 = 2.25

Using tables or computers, we find that the area to the left of this value is equal to 98.78%. This value shows the probability that the length is less than 24 cm, but we ask what is the probability that the length is more than that. We have to subtract this value from 1 (the total area under the curve) and we get 1.2%.

So the probability of exceeding the minimum length is 15.87% and the probability of exceeding the maximum is 1.2%. And we can combine them and say that the probability of exceeding the specified length tolerance is 17.07%.

Is this an academic luxury? Of course not, the examples we have reviewed give important numbers that help the manager to make decisions. In the last example, it seems that the possibility of error is considered great, and therefore this institution either refuses to commit to this work or develops the production method in a significant way that reduces the error rate. In the first example, the restaurant management may find that keeping the drink preparation time to less than 3 minutes in 97.7% of cases is acceptable and may target better than that to reach 99%