# 5.3.1: Practice Using the z Table

- Page ID
- 22051

It's time to practice with the z-table!

Find the \(z\)-score that bounds the *top *9% of the distribution.

**Solution**

Because we are looking for *top *9%, we need to look for the p-value closest to p = .91000 (\(100\% - 9\% = 91\%\)) because the p-values (probabilities) in the z Table show the probability of score being lower, but this question is asking for top 9%, not the portion lower than \(9\%\). There should be \(91\%\) of scores lower than the top \(9\%\).

The closest p-value to p = .91000 (\(91\%\)) is 0.90988. The z-score for p = 0.90988 iz z=1.34.

The z-score for the *top \(*9\%\) of the distribution is z=1.34 (for p=0.90988, the closest probability to \(91\%\), which marks everyone lower than the top \(9\%\)).

Your turn!

Find the \(z\)-score that bounds \(25\%\) (p=0.25000) of the *lower *tail of the distribution.

Hint: You don't have to subtract anything for this one because the question is asking about the scores that are *lower*.

**Answer**-
The z-score for 25% of the lower tail of the distribution is\( z = -0.67\) (for p=0.25143, the closes probability to 0.25000 (\(25\%\))).

Now, let's try some scenarios...

The heights of women in the United States are normally distributed with a mean of 63.7 inches and a standard deviation of 2.7 inches. If you randomly select a woman in the United States, what is the probability that she will be between taller than 64 inches?

**Solution**

X (raw score) = 64 inches

\( \displaystyle \bar{X} \) = 63.7 inches

s=2.7 inches

\[z=\dfrac{x-\overline{X}}{s} = \dfrac{64-63.7}{2.7} = \dfrac{0.30}{2.7} = 0.11 \nonumber \]

Finding z=0.11 on the z Table, we see that p = 0.543860. This is the probability that a score will be lower than our raw score, but the question asked the proportion who would be taller.

\(1 - 0.54380 = 0.4562 \)

\(p \times 100 = 0.4562 \times 100 = 45.62\% \)

Final Answer (in words): The probability that a woman in the U.S. would be 64 inches or taller is 0.4562, or \(45.62\%\).

Your turn!

The heights of men in the United States are normally distributed with a mean of 69.1 inches and a standard deviation of 2.9 inches. What proportion of men are taller than 6 feet (72 inches)?

**Answer**-
Final Answer: The probability that a man in the U.S. would be 72 inches or taller is 0.15866, or \(15.87\%\).

Last one, on something that you might find relevant!

Imagine that you scored 82 points on a final exam. After the final, you find out that the average score on the exam was 78 with a standard deviation of 7. What proportion (in a percentage) did worse than you (earned a *lower *score)?

**Answer**-
The proportion of students in the class who did worse than you (earned a

*lower*score) should be \(71.57\%\).

## Contributors and Attributions

Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)