1. 5.4 Finding Probabilities for a Normal Distribution

2. Continuous Probability Distribution
The histogram and polygon of the distribution in Table 1 is shown to the right.
As stated in Chapter 2, as the class width gets smaller and the number of classes increases, the polygon becomes a smooth curve.
The smooth curve is an approximation of the probability distribution curve of a continuous random variable.

3. Continuous Probability Distribution
A probability distribution of a continuous random variable must satisfy the following two conditions:
The probability that x assumes a value in any of the intervals (class) ranges from 0 to 1. In other words, the area under a probability distribution curve of a continuous random variable between any two points is between 0 and 1.
The total probability of all intervals within which x can assume a value is That is, the total area under the probability distribution curve of a continuous random variable is always 1.0.

4. Properties of a Normal Curve
A normal curve is unimodal and symmetric.
The mean is equal to the median. Both are the center of the curve.

5. Area-Probability Equality Property
The area under a normal curve for an interval is equal to the probability of randomly selecting an observation that lies in the interval.
The total area under a normal curve is equal to 1.

6. Example: Sketching Normal Curves
Suppose that a prestatistics class takes a first test and students’ scores are described well by the normal curve shown in the Figure.

7. Example: Sketching Normal Curves
1. What is the probability that a randomly selected student scored more than 80 points?
2. On Test 2, the mean score is 20 points less than the mean score on Test 1, but the standard deviation is the same. Assuming the scores are still normally distributed, sketch the curve that describes the scores.
3. On Test 3, the mean score is the same as on Test 2, but the standard deviation is larger. Assuming the scores are still normally distributed, sketch a curve that describes the scores.

8. Solution
1. The area to the right of 80, the median, is half of the total area, 1 (see the green region). So, the probability is 0.5.

9. Solution
2. Because the standard deviation hasn’t changed, the shape and spread of the Test 1 and Test 2 distributions are exactly the same. But because the mean is 20 points lower, we move the curve for Test 1 20 points to the left to get the curve for Test 2 (see Figure).

10. Solution
3. Since the means for Test 2 and Test 3 are equal, we do not move the curve for Test 2 left or right (see Fig.). But because the standard deviation is larger for Test 3, we sketch the curve wider. By drawing the curve wider, we must also make it flatter, because the area under the curve must still equal 1. As a check, we could calculate that each little square in the grids of both Test 2 and Test 3 scores and 50 is 0.1 (try it). So, little squares should cover the region below each curve, which is true.

11. Comparing Normal Curves with Different Standard Deviations
A normal curve for a distribution with larger standard deviation will be wider and flatter than a normal curve for a distribution with smaller standard deviation.

12. Empirical Rule
If a distribution is normal, then the probability that a randomly selected observation lies within
one standard deviation of the mean is approximately So, about 68% of the data lie between
two standard deviations of the mean is approximately So, about 95% of the data lie between
three standard deviations of the mean is approximately So, about 99.7% of the data lie between

13. Empirical Rule

14. z-Score
The z-score of an observation is the number of standard deviations that the observation is from the mean. If the observation lies to the left of the mean, then its z-score is negative. If the observation lies to the right of the mean, then its z-score is positive.
z-Score Formula
The z-score of an observation x is given by

15. Example: Compute and Interpret a z-Score
The 2014 draft picks for NBA basketball teams have heights that are approximately normally distributed with mean 79.1 inches and standard deviation inches (Source: nbadraft.net). Shabazz Napier was the shortest 2014 draft pick with a height of 72 inches. Find the z-score for 72 inches. What does it mean?

16. Solution
We substitute 72 for x, 79.1 for , and 3.0 for s in the formula:
The z-score is –2.37, which means that Napier’s height is 2.37 standard deviations less than the mean.

17. z-Scores
Because the z-score of an observation is the number of standard deviations that the observation is from the mean, we can restate the Empirical Rule in terms of z-scores. So, if a distribution is normally distributed, then
68% of its z-scores lie between –1 and 1
95% of its z-scores lie between –2 and 2
99.7% of its z-scores lie between –3 and 3

18. Standard Normal Distribution
If a distribution is normally distributed, then the distribution of the observations’ z-scores is also normally distributed with mean 0 and standard deviation 1. We call the distribution of z-scores the standard normal distribution.

19. Example: Find a Probability for the Standard Normal Distribution
Find the probability that a z-score randomly selected from the standard normal distribution is less than –1.35.
Solution
First, we sketch the standard normal curve, which has mean 0 and standard deviation 1. Next, we mark the z-score –1.35 on the horizontal axis and use orange to shade the region to the left of –1.35.

20. Solution
First, we sketch the standard normal curve, which has mean 0 and standard deviation 1. Next, we mark the z-score –1.35 on the horizontal axis and use orange to shade the region to the left of –1.35.

21. Solution
Then we find the area by using Table 1 in Appendix C. To do this, we write the absolute value of –1.35 as = and identify the row with –1.3 in the far left, identify the column with .05 at the top, and find the number that’s in both that row and column, which is The number describes the area to the left of the z-score –1.35, which is what we are interested in.

22. Example: Using z-Scores and Tables to Find Probabilities
Earlier in this section, we worked with the Wechsler IQ test, which measures a person’s intelligence. Recall that people’s IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Let X be the IQ (in points) of a randomly selected person. Find the probability that a randomly selected person has an IQ
1. less than 78 points.
2. greater than 117 points.

23. Solution
1. We first sketch a normal curve with mean 100, mark the number 78 on the horizontal axis, and use orange to shade the region to the left of 78.
Next calculate the z-score.

24. Solution
Then we find the area by using Table 1 in Appendix C. To do this, we write the absolute value of –1.47 as = and identify the row with –1.4 in the far left, identify the column with .07 at the top, and find the number that’s in both that row and column, which is The number describes the area of the orange region to the left of 78 points, which is what we are interested in. So, the probability is The probability is less than 0.5, which checks with the orange region having area less than half of the area under the entire curve.

25. Solution
2. We first sketch a normal curve with mean 100, mark the number 117 on the horizontal axis, and use blue to shade the region to the right of 117.

26. Solution
Then we use Table 1 in Appendix C to find an area. To do this, we write 1.13 = and identify the row with 1.1 in the far left, identify the column with at the top, and find the number that’s in both that row and column, which is The number describes the area (in orange) to the left of 117 points, which is not what we are interested in. Because the area under the entire curve is 1, we can find the area (in blue) to the right of 117 by finding 1 minus : 1 – =

27. Solution
So, the probability is , which we indicate in the figure. The probability is less than 0.5, which checks with the blue region having area less than half of the area under the entire curve.

28. Interpreting the Area under a Normal Curve for an Interval
We can interpret the area A under the normal curve for an interval as
the probability that a randomly selected observation will lie in the interval.
the proportion of observations that lie in the interval.