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Example 6.8 from Statistics, 6.2 Using the Normal Distribution
(part 1)
a. Find the probability that a randomly selected student scored more than 65 on the exam.
Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.
Solution
a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5.
Draw a graph.
Calculate the z-score:
6.4The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446.
Then, find P(x > 65).
Figure 6.5The probability that any student selected at random scores more than 65 is 0.3446.
Using the TI-83, 83+, 84, 84+ Calculator
Go into
2nd DISTR.
After pressing2nd DISTR, press2:normalcdf.The syntax for the instructions is as follows:
normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing
1, theEEkey—a 2nd key—and then99. Or, you can enter10^99instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose.Historical Note
The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table, also known as a Z-table—the math involved to find probability is cumbersome. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right.
How did it go?