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Normal Probabilities for AP Statistics

By — McGraw-Hill Professional
Updated on Feb 3, 2011

Practice problems for these concepts can be found at:

When we know a distribution is approximately normal, we can solve many types of problems.

    example: In a standard normal distribution, what is the probability that z < 1.5? (Note that because z is a CRV, P(X = a) = 0, so this problem could have been equivalently stated "what is the probability that z ≤ 1.5?")
    solution: The standard normal table gives areas to the left of a specified z-score. From the table, we determine that the area to the left of z = 1.5 is 0.9332. That is, P(z < 1.5) = 0.9332. This can be visualized as follows:

Normal Probabilities

    example: It is known that the heights (X) of students at Downtown College are approximately normally distributed with a mean of 68 inches and a standard deviation of 3 inches. That is, X has N(68,3). Determine
  1. P(X < 65).
  2. solution: P(X < 65) = = 0.1587 (the area to the left of z = –1 from Table A). On the TI-83/84, the corresponding calculation is normalcdf (–100,-1) = normalcdf(–1000,65,68,30) = 0.1586552596.

  3. P(X > 65).
  4. solution: From part (a) of the example, we have P(X < 65) = 0.1587. Hence, P(X > 65) = 1 – P(X < 65) = 1 – 0.1587 = 0.8413. On the TI-83/84, the corresponding calculation is normalcdf(–1,100)= normalcdf(65,1000,68,3)= 0.8413447404.

  5. P(65 < X < 70).
  6. solution: P(65 < X < 70) = 0.7486 – 0.1587 = 0.5899 (from Table A, the geometry of the situation dictates that the area to the left of z = –1 must be subtracted from the area to the left of z = 0.667). Using the TI-83/84, the calculation is normalcdf (–1,0.67) = normalcdf (65,70,68,3)=0.5889. This situation is pictured below.

    Normal Probabilities

    Note that there is some rounding error when using Table A (see Appendix).

    In part (c), z = 0.66667, but we must use 0.67 to use the table.

  7. P(70 < X < 75)
  8. solution: Now we need the area between 70 and 75. The geometry of the situation dictates that we subtract the area to the left of 70 from the area to the left of 75. This is pictured below

    Normal Probabilities

    We saw from part (c) that the area to the left of 70 is 0.7486. In a similar fashion, we find that the area to the left of 75 is 0.9901 (based on z = 2.33). Thus P(70 < X < 75) = 0.9901 – 0.7486 = 0.2415. The calculation on the TI-83/84 is: normalcdf (70,75,68,3)= 0.2427. The difference in the answers is due to rounding.

    example: SAT scores are approximately normally distributed with a mean of about 500 and a standard deviation of 100. Laurie needs to be in the top 15% on the SAT in order to ensure her acceptance by Giant U. What is the minimum score she must earn to be able to start packing her bags for college?
    solution: This is somewhat different from the previous examples. Up until now, we have been given, or have figured out, a z-score, and have needed to determine an area. Now, we are given an area and are asked to determine a particular score. If we are using the table of normal probabilities, it is a situation in which we must read from inside the table out to the z-scores rather than from the outside in. If the particular value of X we are looking for is the lower bound for the top 15% of scores, then there are 85% of the scores to the left of x. We look through the table and find the closest entry to 0.8500 and determine it to be 0.8508. This corresponds to a z-score of 1.04. Another way to write the z-score of the desired value of X is

Normal Probabilities.

    Thus, = 1.04.

Solving for x, we get x = 500 + 1.04(100) = 604. So, Laurie must achieve an SAT score of at least 604. This problem can be done on the calculator as follows: invNorm(0.85,500,100).

Most problems of this type can be solved in the same way: express the z-score of the desired value in two different ways (from the definition; finding the actual value from Table A or by using the invNorm function on the calculator), then equate the expressions and solve for x.

Practice problems for these concepts can be found at:

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