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PreMBA Analytical Methods
Continuous Probability Distributions: Standard Normal Distribution

Suppose you have a normally distributed random variable and would like to calculate the probability of its value occurring in the interval of the mean plus or minus .5 standard deviations. How would you go about calculating the probability?

One method is to standardize the random variable. Standardization of a normally distributed random variable enables an analyst or researcher to determine with ease the probability associated with a range of values for that variable by using a standardized distribution table. A normally distributed random variable can be standardized using a formula. The letter Z represents the standardized random variable and the probabilities associated with ranges of values of Z can be found in a Z distribution table. Read on to learn more about how to calcultate (and apply) the standardized normally distributed random variable Z.

Standard Normal Distribution

The standardized value of a normally distributed random variable is called a Z score and is calculated using the following formula.

x = the value that is being standardized
m = the mean of the distribution
s = standard deviation of the distribution

As the formula shows, a random variable is standardized by subtracting the mean of the distribution from the value being standardized, and then dividing this difference by the standard deviation of the distribution. Once standardized, a normally distributed random variable has a mean of zero and a standard deviation of one. The standard normal distribution (Z distribution) is shown in the graph below. As you can see from the notation to the right of the curve, mz = 0 and sz = 1.

Learn How to Calculate the Mean of a Standard Normal Distribution

To see how a random variable is standardized, imagine you have a random variable, x, that is normally distributed with a mean of 20 and a standard deviation of 10. What would be the standardized value (Z score) of 40? To solve this problem, you would use the Z score formula. Using the information from the example, you know that

x = 40
m = 20
s = 10

Substitute these numbers into the formula and solve.

In this example, the standardized value (Z score) of 40 is 2.

These Z scores are important because they tell you how far a value is from the mean. When you standardize a random variable, its mean becomes zero and its standard deviation becomes one; therefore,

  • if the Z score of x is zero, then the value of x is equal to the mean.

  • if the Z score of x is one, then the value of x is one standard deviation above the mean. If the Z score is –1, then the value of x is one standard deviation below the mean.

  • If the Z score of x is two, then the value of x is two standard deviations above the mean. If the Z score is –2, then the value of x is two standard deviations below the mean.
  • In the earlier example, the Z score of 40 was 2. This tells you that 40 is 2 standard deviations above the mean. Consider three other values from the same distribution and their Z scores, shown in the table below.

    As the table shows, the value of 10 has a Z score of –1, which means it is one standard deviation below the mean. The value of 20 has a Z score of 0, which means it is equal to the mean. And the value of 30 has a Z score of 1, which means it is one standard deviation above the mean.

    Calculating Probability Within a Desired Range

    The Z score allows you to calculate the probability that a normally distributed random variable x will fall within a desired range. As an example, consider an apparel company that produces garments in China and then ships them to the United States for and sale.

    Let the random variable x be the time required to transport the product from the manufacturing facility in China to the United States. Suppose that x is normally distributed with a mean of 24 days and a standard deviation of 3 days. What is the probability that a shipment will arrive within 30 days—P(x less than or equal to 30)?

    To determine the probability, you must first calculate the Z score. With the information from the example, you know that

    x = 30
    m = 24
    s = 3

    Substitute this information into the Z score formula, and standardize the value of 30, as shown below.

    Learn How to Calculate the Standard Deviation of a Standard Normal Distribution

    The next step in determining the probability that the shipment will arrive within 30 days is to use a Z distribution table. This table will tell you the probability associated with a given Z score.

    Z Distribution Table

    The Z distribution is a normal distribution that looks much like the graph below. The figures in the table show the probability that a random variable x will fall within the range to the left of the Z score (from negative infinity up to the Z score.) Consider the diagram below for a more intuitive understanding of this concept. If you looked up a Z score of two in a Z distribution table, you would find that the probability listed in the table represents the pink shaded area under the curve in the illustration. If you are interested in the probability of the clear area under the curve—P(Z > 2)—you would have to subtract the probability found in the table from one to determine the probability of Z greater than two.

    As you might expect, a Z score of zero, when looked up in the table, yields a probability of .5. This makes sense if you remember the definition of a normal distribution having half of its values distributed on either side of the mean.

    Returning to the example of apparel shipments, you can look up the Z score of 2.00 in the table to find the probability that the shipment will arrive within 30 days. The probability can be found by looking first down the left-hand column of the table for the number 2.0. Since the number after the decimal point is zero, use the first column titled .00 for the probability. The probability that Z will be less than or equal to 2.0 is .9772. (This number is highlighted in this Z table.) This means that the probability that the shipment will arrive within 30 days is .9772.

    With any normally distributed random variable, you can find the probability of the variable x falling between two values: calculate the Z score of each value, look up their probabilities in a Z distribution table, and then subtract the smaller probability from the larger one. The difference is the probability of the random variable x falling between the two values.

    For a demonstration of finding the probability of an interval of values, consider again the apparel manufacturer example from above. Calculate the probability of an apparel shipment arriving within the range of 22 to 26 days, using above data for mx and sx of 24 days and 3 days, respectively.

    To solve this problem, first standardize the variables at each end of the interval in question. Calculate the Z value using the following formula for Z.

    First, for the lower end of the interval of 22 days, calculate the Z value.

    Then, for the upper end of the interval of 26 days, calculate the Z value.

    Now look up the probability of each standardized random variable Z in a Z distribution table. The probability associated with the standardized random variable –.67 is .2514 (a value which is determined by calculating 1 – .7486). The probability associated with the standardized random variable .67 is .7486.

    To find the probability of an interval, subtract the lower end of the interval from the upper end of the interval.

    The probability of an apparel shipment arriving in 22 to 26 days is .4972.

    1. What is the assumed distribution behind a standardized random variable, Z?

    Solution 1

    2. Why is the concept of a standardized variable important?

    Solution 2

    3. What is the probability that a standardized random variable Z is less than or equal to 1.96—P(Z < 1.96)?

    Solution 3

    4. What is the probability that a Z score is greater than or equal to 1.5 and less than or equal to 2.0—
    P(1.5 < Z < 2.0)?

    Solution 4

    5. If the population of monthly cell phone bills of a group of students has a mean of $43.00 (m = $43.00) and a standard deviation of $5.70 (s = $5.70), what is the probability of selecting a student at random whose monthly bill is $36.50 or less?

    Solution 5

    6. The manufacturer of a box of cookies states on the outside of each box that the product contents weigh 16 ounces. The population of boxes of cookies has a mean of 16.25 ounces (m = 16.25 ounces) and a standard deviation of .30 ounces (s = .30 ounces).

    The manufacturer would like to ensure that the contents of each box weigh between 15.90 and 16.90 ounces. Less than 15.90 ounces may lead to customer satisfaction issues, while more than 16.90 ounces results in too much product being given away in each box. What is the probability of selecting a box of cookies with product content weighing between 15.90 and 16.90 ounces?

    Solution 6

    How To Read a Z Distribution Table
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