PSCY 510 Data Organization & Descriptive Statistics

PSYC 510 WORKSHEET: Week 4 presentation “z scores, normal curve, and probability”

Review of Material: z scores and probability (Week 4)

Information from: Jackson Ch. 5 section “ z-Scores” and Week 4 presentation “z scores, normal curve, and probability”

Additional practice: Jackson ch. 5 chapter exercise #7A – F (answers in e-book for self-check)

Note: this is an OPTIONAL worksheet to practice applying some of this week’s key concepts. Please make sure you complete all assigned readings and watch this week’s presentations before attempting the worksheet! Try to complete it on your own, then check your answers with the answer key details (at the end of the document).

Scenario: The following anxiety subscale data is for teachers based on the ECR-RT questionnaire (ECR-RT = “Experiences in close relationships-revised for teachers) (see Table 1 in Riley, 2013). Assume for the purposes of this practice that the distribution for each of the samples below is normal.

For teachers between the ages of 20 – 24, the mean = 53.80 and the standard deviation = 18.51.

For teachers 35 years of age and older, the mean = 42.18 and the standard deviation = 13.01.

Reference: Riley, Philip. (2013). Attachment theory, teacher motivation & pastoral care: A challenge for teachers and academics. Pastoral Care in Education, 31 (2). DOI: 10.1080/02643944.2013.774043.

Practice Problems

1. Juan, a male teacher aged 23, scored 65 on the ECR-RT subscale for anxiety. What proportion of people in his age group would score equal or higher than him?

a. What is Juan’s percentile rank?

2. What proportion of teachers aged 20 – 24 would score 50 or lower?

3. What score would be at the 60th percentile for teachers over the age of 35 years?

4. What score would be at the 25th percentile for teachers over the age of 35 years?

5. What is the likelihood that a 42 year old teacher would either up to 33.46 OR higher than 45.43?

6. What is the likelihood that a 42 year old teacher would score between a 40 and 50?

ANSWER KEY z scores, normal curve, and probability (Week 4)

Remember, z-scores can be used for any normally distributed data in order to standardize its comparison, as it removes the units of measurement. As discussed this week, the normal curve has a mean of 0 and a standard deviation of 1. Statisticians have calculated all the probabilities at any given area on this curve, and these values are provided in Appendix A of your e-book. So if you are provided a mean and standard deviation, you can use this information to discuss where a raw score would fall using z scores.

1. Juan, a male teacher aged 23, scored 65 on the ECR-RT subscale for anxiety. What proportion of people in his age group would score equal or higher than him?

The formula is:

Juan is 23 so we will use the sample group for teachers 20 – 24. For this group, the provided mean = 53.8, and standard deviation S = 18.51. “X” denotes the individual’s value, which is 65. Thus, we plug these numbers into the formula and get: = 0.605078336. This is the z score. I like to think about where this falls on the normal curve. Here’s our normal curve, and I’ll put where the (positive) 0.605078336 would fall. The Normal Curve is in Appendix A.2, like shown above. We know the z value (0.61 since we round to two decimal places, as that is all that is provided for the z scores. Anything ending in the third decimal with a 5 or higher would round up; 4 or lower rounds down). From the table, we see:

z 0.61

Area between mean and z 0.22908

Area beyond z 0.27092

The question asks what proportion of people would score equal or higher than him. So, we’d want to know the area shaded with red. This is the area beyond the z, and is 0.27092 – your final answer.

a. What is Juan’s percentile rank?

There is several ways to answer this – whatever works best for you! The first is knowing the entire area under the curve = 100%, and that in the question above, 0.27092 proportion = 27.092%. Thus, we’d just need to subtract 100 – 27.092 to get to where Juan is on the normal curve. ANSWER: 72.908%. To double check or not rely on previous answers, we would solve for z as we did above and get 0.61. But, we’d use the area between the mean and z to get 0.22908 – but wait we aren’t done!! That only covers from the midline to 0.6. We need to add the other half of the normal curve. Knowing the total is 1.0, half would be 0.5. So, you add 0.22908 + 0.5 = 72.908%

2. What proportion of teachers aged 20 – 24 would score 50 or lower?

In this question, you are again given the “X”. In our scenario we are given the mean and standard deviation. So, again we simply plug the numbers into the formula: = -0.20529 (round to -0.21). However – look at where this falls on the normal curve (shown in purple). We should be able to have guessed this, given the 50 is lower than the mean (the mean would fall directly on the center).

z 0.21

Area between mean and z 0.08318

Area beyond z 0.41682

This question asks 50 or LOWER, so I’ve shaded in what would answer this question (lower / less is always to the left; higher / more is always to the right). So, to answer this question we need the numbers in the table from “area beyond the z”. Our answer i s 0.41682 . (note if the question asked for what PERCENT, we’d merely multiple this by 100, so it’d be 41.682%).

3. What score would be at the 60th percentile for teachers over the age of 35 years?

Now we are using the other mean and standard deviation provided in the scenario: = 42.18; S = 13.01. However, we are NOT given the X for our equation, nor are we given a z. However – we can use our table to calculate our z! Knowing the entire area under the curve is 100%, where would 60% fall? On the right.. We know our z table only shows half the values for the tables, so there won’t be anything above 0.5 (50%). Thus, we need to find a column in our z table where we see the closest thing to 10%. In proportions, 10% = 0.10 (divide the percent by 100). Which column should we read though? Think about where 60% falls on this normal curve. It’d be on the right-hand side, so getting up to the 60th percent, we need to go through the table in the “area between mean and z” to find the one closest to 0.10. These two are pretty close but which one is closest to 0.10? Sign doesn’t matter – just which one is closest. After subtracting both from .10, we see that the z score of 0.25 is closest.

z	Area between mean and z	Area beyond z
0.25	0.09872 (-0.10 = .00128* smaller than .00257)	0.40128
0.26	0.10257 (-0.10 = .00257)	0.39743

We have z, mean, and standard deviation. Now we have to solve for X. You can transform the equation yourself, or see it transformed in the chapter to be: (note I switched out the population for sample symbols, since we are talking about samples here but they are interchangeable in this context).

Filling out what is provided, we get: = 45.4325. So, the 60th percentile score expected for this group of teachers aged 35 or older is 45.4325.

4. What score would be at the 25th percentile for teachers over the age of 35 years?

Using the same mean and standard deviation as provided for this age group, we have = 42.18; S = 13.01. However, we are not explicitly given the z score or the X. But, we can use our table to determine z. The 25th percentile would fall on which side? The left . So, we will need to read the column “area beyond z” to find what z most closely corresponds to 25th percent (which would be 0.25 proportion). Again, there are two fairly close. Subtracting each from .25 reveals that the z score of 0.67 is closest. But WAIT!!

z	Area between mean and z	Area beyond z
0.67	0.24858	0.25142 (-.25 = .00142* smaller than .00175)
0.68	0.25175	0.24825 (-.25 = .00175)

Remember, we are on the LEFT side so z does NOT equal 0.67. It is -0.67 (NEGATIVE – see normal curve below). This is crucial!! Thus, using , we get = 33.4633 is our final answer. A good way to check yourself is knowing a percentile rank question with the percentile below 50% should end with a value below the average. A percentile rank question with the percent above 50% should end with a value above the mean.

5. What is the likelihood that a 42-year-old teacher would score either up to 33.46 OR higher than 45.43?

First, solve for z for both scores.

= -0.67 and = 0.25

The good thing about this question is we have the probabilities from questions 3 and 4 (purposeful, since these are practice and I figured your time is precious ). So, we have our charts we just have to think about which column to read from for each likelihood.

· “Up to 33.46” (z score of -0.67) would shade in from the far left of a normal curve to -0.67, so we care about the proportion in the “area beyond the z”, or 0.25142.

· “Higher than 45.43” would shade the far right of a normal curve from 0.25 on, so it’d also use the proportion in the “area beyond the z” but for 0.25, which is 0.40128

· We want to know the likelihood of one or the other. This is a probability question and since it is an “OR” question, you use the addition rule (see chapter 7). Thus, we just need to add our two probabilities together.

· 0.25142 + 0.40128 = 0.6527 – the probability a 42 year old teacher will score either up to 33.46 or higher than 45.43.

6. What is the likelihood that a 42-year-old teacher would score between a 40 and 50?

First, we need to calculate our z scores.

= -0.17 and = 0.60

Now, we need to think about what the question is asking to “shade in” what we are looking for so that when we consult our charts we can know which column to read.

z	Area between mean and z	Area beyond z
0.17	0.06751	0.43249
0.60	0.22575	0.27425

For the -0.17 we need the “area between the mean and z”, or 0.06751.

-0.17

0.60

· For our z score of 0.60, we need the proportion in the “area between the mean and z”, which is 0.22575.

· We want to know the likelihood of this range so we add them together. 0.06751 + 0.22575 = 0.29326

· We can state that the likelihood that a 42 y/o teacher will score between 40 – 50 is 0.29326.

NOTE: There are still opportunities for extra practice! Try completing Jackson ch. 5 chapter exercise #7A – F (answers in e-book for self-check)