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)
Standard Deviation (σ)
Probability (Area)
Exercise of contract performanceA 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.


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 σ2; that is, X ~ N(μ,σ2), which will become the standard normal random variable, Z ~ N(0,1). To do so, we will apply the following transformation:

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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