# Exercise of contract performanceInverse Normal Distribution Example

### Problem Data

It is required to calculate the limit of the random variable X given the **left tail probability** taking into consideration the following data:

- Mean (μ)
- 500000
- Standard Deviation (σ)
- 50000
- Probability (Area)
- 0.2
- Description
**Exercise of contract performance**A contractor considers the cost of fulfilling a contract to be a random variable following a normal distribution having a mean of $500,000 and a standard deviation of $50,000. For up to what level of cost does one have a probability of fulfilling the contract of 0.2.

## Solution

Below are the detailed calculations and graphs to obtain **the limit** according to the data provided:

### Step 1:

We have the normal random variable X with mean μ and variance σ^{2}; that is, ** X ~ N(μ,σ^{2})**, which will become the standard normal random variable,

**. To do so, we will apply the following transformation:**

*Z ~ N(0,1)*We have μ = 500000 and σ = 50000, then:

Given the left tail boundary * b*, we have the following:

Once the expression has been transformed to a normal distribution, it is necessary to know for which value of Z this probability is obtained. Performing computer calculations, the exact value of Z to obtain a right-tailed probability of 0.2 is **-0.8416**.

Using this value, we replace the result in the previous equation:

**Calculation with table**

You can also use your statistical table to solve the problem, however the result may not be exact (it will be an approximation), because the statistical table only considers Z values with two decimal places. The procedure will be as follows:

**Adapt the expression to the values in the table:**In our statistical table the probabilities shown range from 0.5 to 1. Therefore we can apply the following expression:**Search in the table matrix:**In the table results, find the number closest to 0.8 and mark it. In our table the closest number is 0.7995.- From the marked number, we locate the values of its corresponding
**row=0.8**and**column=0.04**. Finally, we add both values and obtain the result**0.84**

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