Areas Under the Normal Distribution Curve

Figure 1. Normal Distribution Table

Finding Areas Under the Normal Distribution Curve

          In finding areas under the normal curve, used table in Figure 1. This table provides areas under the standard normal curve below a specified value z. The first column in table of normal curve refers to the values of the standard score or z-score ranging from 0.0 to +3.0 standard deviations from the mean. The first-row heading specifies the value of z-score in two decimal places while the entries in the table are the areas below a given value of z, in four-decimal-place accuracy.

Example 1: Determine the area under the standard normal curve of z = 2.49.

Solution: First go down the left-hand column, labeled z to “2.4”. Then go cross that row until the column 0.09. That intersection is the area under the standard normal curve of z = 2.49, that is, 0.4936.

Example 2: Determine the area under the standard normal curve of z = -1.38.

Solution: Apply the property of the normal curve that is, the normal curve is symmetrical about the mean. Thus, the area under the normal curve of z = -1.38 is just the same in finding areas under the normal curve of z = 1.38. First go down the left-hand column, labeled z to “1.3”. Then go cross that row until the column 0.08. The intersection is the area under the standard normal curve of z = -1.38, that is, 0.4162.

Example 3: Find the area under the standard normal curve to the right of z = 0.25.

Solution: First go down the left-hand column, labeled z to “0.2”. Then go cross that row until the column 0.05. That intersection is the area under the standard normal curve from 0 to z = 0.25 which is 0.0987, thus, the area under the standard normal curve to the right of z = 0.25 is the difference between 0.5 and obtained area or 0.5 – 0.0987 = 0.4013.

Example 4: Find the area under the standard normal curve to the left of z = 1.73.

Solution: First go down the left-hand column, labeled z to “1.7”. Then go cross that row until the column 0.03. That intersection is the area under the standard normal curve from 0 to z = 1.73 which is 0.4582, thus, the area under the standard normal curve to the left of z = 1.73 is the sum between 0.5 and obtained area or 0.5 + 0.4582 = 0.9582.

Example 5: Find the area under the standard normal curve to the left of z = -1.08.

Solution: First go down the left-hand column, labeled z to “1.0”. Then go cross that row until the column 0.08. That intersection is the area under the standard normal curve from 0 to z = -1.08 which is 0.3599, thus, the area under the standard normal curve to the left of z = -1.08 is the difference between 0.5 and obtained area or 0.5 – 0.3599 = 0.1401.

Example 6: Find the area under the standard normal curve above z = -2.46.

Solution: First go down the left-hand column, labeled z to “2.4”. Then go cross that row until the column 0.06. That intersection is the area under the standard normal curve from 0 to z = -2.46 which is 0.4931, thus, the area under the standard normal curve to the left of z = -2.46 is the sum between 0.5 and obtained area or 0.5 + 0.4931 = 0.9931.

Example 7: Find the area under the standard normal curve between z = 0.82 and z = 2.72.

Solution: The area between 0 and 0.82 is 0.2939 and the area between 0 and 2.72 is 0.4967. Thus, the required area between the z values 0.82 and 2.72 is the difference between these two areas or 0.4967 – 0.2939 = 0.2028.

Example 8: Determine the area under the standard normal curve between z = -1.57 and z = 1.63.

Solution: The area between -1.57 and 0 is 0.4418 and the area between 0 and 1.63 is 0.4484. Thus, the required area between the z values -1.57 and 1.63 is the sum between these two areas or 0.4484 – 0.4418 = 0.8902.

Standardizing a Normal Curve

          Most of the time, the probabilities you are interested in will involve a random variable let say x, a normal random variable with mean μ and standard deviation σ. You must then standardize variable of interest, writing it as the equivalent form in terms of z-score the standard normal random variable. Once this is done, the probability of interest is the area that you find using the standard normal probability distribution. The formula to convert raw score into standard score is:

Example 1: For a continuous random variable that has a normal distribution with a mean of 20 and a standard deviation of 4, find the area under the normal curve from 19 and 26.

Now, the area between -0.25 and 0 is 0.0987 and the area between 0 and 1.5 is 0.3531. Therefore, the area under the normal curve between 19 and 26 is sum between these two areas or 0.3531 + 0.0987 = 0.4518.

Example 3: A brisk walk at 4 miles per hour burns an average of 300 calories per hour. If the standard deviation of the distribution is 8 calories, find the probability that a person who walks one hour at the rate of 4 miles per hour will burn the following calories. Assume the variable to be randomly normally distributed.

a. more than 280 calories.

b. less than 294 calories.

c. between 278 and 318 calories.

The area from -0.75 and 0 is 0.2734. Thus, the probability that a person who walks one hour at the rate of 4 miles per hour will burn less than 294 calories is the difference between 0.5 and the area or 0.5 – 0.2734 = 0.2266. Therefore, there is a 22.66% probability that a person who walks one hour at the rate of 4 miles per hour will burn less than 294 calories.

The area between -2.75 and 0 is 0.4970 and the area between 0 and 2.25 is 0.4878., the probability is the sum between these two areas or 0.4878 + 0.4970 = 0.9848. Therefore, around 98.48% probability that a person who walks one hour at the rate of 4 miles per hour will burn between 278 and 318 calories respectively.

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