Homework 4 Normal Curve And Z Scores Statistics
PSYC 354 H OMEWORK 5 ZScores Be sure you have reviewed this module/week’s lessons and presentations along with the practice data analysis before proceeding to the homework exercises. Complete all analyses in SPSS, and then copy and paste your output and graphs into your homework document file. Number all responses. Answer any written questions (such as the textbased questions or the APA Participants section) in the appropriate place within the same file. Review the “Homework Instructions: General” document for an example of how homework assignments must look. Part I: Concepts These questions are based on the Nolan and Heinzen reading and endofchapter questions. 1. What are always the mean and standard deviation of the zdistribution? 0 and 1 2. Define the central limit theorem. The central limit theorem refers to how a distribution of sample means is a more normal distribution than a distribution of scores, even when the population distribution is not normal. 3. Fill in the blanks: A zscore is based on a distribution of _____ standard score ________, while a zstatistic is based on a distribution of _______ mean ___________. 4. Endofchapter problems: Remember to show work to receive partial credit where applicable. For help working on these problems, refer to the presentation from this module/week on the normal curve and computing zscores. Raw and zscores: 6.16 and 6.20 6.16 If a population has a mean of 250 and a standard deviation of 47, calculate z scores for each of the following raw scores: a. 391 z = (391250)/47 = 3.0 b. 273 z = (273250)/47 = 0.49 c. 199 z = (199250)/47 = 1.09 d. 160 z = (160250)/47 = 1.91 6.20 For a population with a mean of 250 and a standard deviation of 47, convert each of the following z scores to raw scores. a. 0.54 x = .54(47) + 250 = 275.38 b. 2.66 x = 2.66(47) +250 = 124.98 c. 1.0 x = 1.0(47) +250 = 203 Page 1 of 9

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, (see your Course Schedule).
Probability Distribution for Continuous Random Variable
A continuous random variable is a variable that assumes values in an interval.
We define the probability distribution of Y as f(y) where: P(a < Y < b) is the area under f(y) over the interval from a to b.
Note: If Y is continuous P(Y = y) = 0 for any given value y.
Normal Distribution with Mean \(\mu\), Standard Deviation \(\sigma\)
Normal distribution is a family of continuous distributions that can model many histograms of reallife data which are moundshape and symmetric (for example, height, weight, etc.).
A normal curve has two parameters:
 mean \(\mu\) (center of the curve)
 standard deviation \(\sigma\) (spread about the center)
In particular, a normal distribution with \(\mu\) = 0 and \(\sigma\) = 1 is called a standard normal curve, usually denoted as Z. Most statistics books provide tables to display the area under a standard normal curve. Using them, one can calculate area under a general normal curves. Look in the appendix of your textbook for the Standard Normal Table. We include a similar table, Standard Normal Cumulative Probability Table here so that you can print the table out and refer to it easily when working on the homework. In the following examples, we use this Standard Normal Cumulative Probability Table to obtain the needed values.
Finding Probability Under the Standard Normal Curve Given the Zvalues
Find the area under the standard normal curve to the right of 0.87.
Answer:
P(Z > 0.87)
= 1  P(Z ≤ 0.87)
= 1  0.8078
= 0.1922
A typical fourdecimalplace number in the body of the Standard Normal Cumulative Probability Table gives the area under the standard normal curve that lies to the left of a specified zvalue. The probability to the left of z = 0.87 is 0.8078 and it can be found by reading the table:
 Since z = 0.87 is positive, use the table for POSITIVE zvalues.
 First, go down the lefthand column, label z to "0.8."
 Then, go across that row until under the "0.07" in the top row.
z  .00  .01  .02  .03  .04  .05  .06  .07  .08  .09 
0.0  .5000  .5040  .5080  .5120  .5160  .5199  .5239  .5279  .5319  .5359 
0.1  .5398  .5438  .5478  .5517  .5557  .5596  .5636  .5675  .5714  .5753 
0.2  .5793  .5832  .5871  .5910  .5948  .5987  .6026  .6064  .6103  .6141 
0.3  .6179  .6217  .6255  .6293  .6331  .6368  .6406  .6443  .6480  .6517 
0.4  .6554  .6591  .6628  .6664  .6700  .6736  .6772  .6808  .6844  .6879 
0.5  .6915  .6950  .6985  .7019  .7054  .7088  .7123  .7157  .7190  .7224 
0.6  .7257  .7291  .7324  .7357  .7389  .7422  .7454  .7486  .7517  .7549 
0.7  .7580  .7611  .7642  .7673  .7704  .7734  .7764  .7794  .7823  .7852 
0.8  .7881  .7910  .7939  .7967  .7995  .8023  .8051  .8078  .8106  .8133 
0.9  .8159  .8186  .8212  .8238  .8264  .8289  .8315  .8340  .8365  .8389 
1.0  .8413  .8438  .8461  .8485  .8508  .8531  .8554  .8577  .8599  .8621 
Find the area under the standard normal curve between 2 and 3.
Answer:
P(2 < Z < 3)
= P(Z < 3)  P(Z ≤ 2)
= 0.9987  0.9772
= 0.0215
The entry in the Standard Normal Cumulative Table corresponding to z = 3.0 is 0.9987 and to z = 2.0 is 0.9772.
z  .00  .01  .02  .03  .04  .05  .06  .07  .08  .09 
.  .  .  .  .  .  .  .  .  .  . 
.  .  .  .  .  .  .  .  .  .  . 
2.0  .9772  .9778  .9783  .9788  9793  .9798  .9803  .9808  .9812  .9817 
2.1  .9821  .9826  .9830  .9834  .9838  .9842  .9846  .9850  .9854  .9857 
2.2  .9861  .9864  .9868  .9871  .9875  .9878  .9881  .9884  .9887  .9890 
2.3  .9893  .9896  .9898  .9901  .9904  .9906  .9909  .9911  .9913  .9916 
2.4  .9918  .9920  .9922  .9925  .9927  .9929  .9931  .9932  .9934  .9936 
2.5  9938  .9940  .9941  .9943  .9945  .9946  .9948  .9949  .9951  .9952 
2.6  .9953  .9955  .9956  .9957  .9959  .9960  .9961  .9962  .9963  .9964 
2.7  .9965  .9966  .9967  .9968  .9969  .9970  .9971  .9972  .9973  .9974 
2.8  .9974  .9975  .9976  .9977  .9977  .9978  .9979  .9979  .9980  .9981 
2.9  .9981  .9982  .9982  .9983  .9984  .9984  .9985  .9985  .9986  .9986 
3.0  .9987  .9987  .9987  .9988  .9988  .9989  .9989  .9989  .9990  .9990 
The above two examples are to find area (probabilities) given the zvalue. Next, we will show how to find the zvalue given the area (probabilities).
Finding the Zvalues Given the Probability
Find the 10th percentile of the standard normal curve. They are asking for a zvalue to the left of which has an area of 0.1 under the standard normal curve.
Answer: Since the entries in the Standard Normal Cumulative Probability Table represent the probabilities and they are fourdecimalplace numbers, we shall write 0.1 as 0.1000 to remind ourselves that it corresponds to the inside entry of the table. We search the body of the tables and find that the closest value to 0.1000 is 0.1003. We look to the leftmost of the row and up to the top of the column to find the corresponding zvalue.
The corresponding zvalue is 1.28. Thus z = 1.28.
z  .00  .01  .02  .03  .04  .05  .06  .07  .08  .09 
.  .  .  .  .  .  .  .  .  .  . 
.  .  .  .  .  .  .  .  .  .  . 
1.3  .0968  .0951  .0934  .0918  .0901  .0885  .0869  .0853  .0838  .0823 
1.2  .1151  .1131  .1112  .1093  .1075  .1056  .1038  .1020  .1003  .0985 
1.1  .1357  .1335  .1314  .1292  .1271  .1251  .1230  .1210  .1190  .1170 
Here is another explanation, this time walked through in a movie type format. VIEWLET: How to use standard normal table to find probability to the left of z.
Find the zvalue having an area of 0.2 in the upper tail. Try to figure out your answer first, then click the graphic to compare answers.
Find the zvalue having an area of 0.2 in the upper tail.
Answer:
This is the same as asking for the zvalue with area 0.8000 to its left. The value in the Standard Normal Cumulative Probability Table that is closest to 0.8000 is 0.7995.
The corresponding zvalue is 0.84. Thus, z = 0.84
z  .00  .01  .02  .03  .04  .05  .06  .07  .08  .09 
0.0  .5000  .5040  .5080  .5120  .5160  .5199  .5239  .5279  .5319  .5359 
0.1  .5398  .5438  .5478  .5517  .5557  .5596  .5636  .5675  .5714  .5753 
0.2  .5793  .5832  .5871  .5910  .5948  .5987  .6026  .6064  .6103  .6141 
0.3  .6179  .6217  .6255  .6293  .6331  .6368  .6406  .6443  .6480  .6517 
0.4  .6554  .6591  .6628  .6664  .6700  .6736  .6772  .6808  .6844  .6879 
0.5  .6915  .6950  .6985  .7019  .7054  .7088  .7123  .7157  .7190  .7224 
0.6  .7257  .7291  .7324  .7357  .7389  .7422  .7454  .7486  .7517  .7549 
0.7  .7580  .7611  .7642  .7673  .7704  .7734  .7764  .7794  .7823  .7852 
0.8  .7881  .7910  .7939  .7967  .7995  .8023  .8051  .8078  .8106  .8133 
What about if we had a problem that was the other way around? We know the probability and we want to find the zvalue? Walk through how this is done: VIEWLET: Given probability, how to find z value.
Using the Standard Normal Table for Normal Data
Remember, the table can be used with normal data that has been standardized. This allows us to compare observations from different normal random variables. For instance, assume U.S. adult heights and weights are both normally distributed. Clearly, they would have different means and standard deviations. However, if you knew these means and standard deviations, you could find your zscore for your weight and height by recalling the zscore formula that we used Lesson 2 where \(z = \frac{\text{observed}\mu}{\sigma}\).
You can now use the Standard Normal Table to find the probability, say, of a randomly selected U.S. adult weighing less than you or taller than you.
According to the Center of Disease Control, heights for U.S. adult females and males are approximately normal.
Females: mean of 64 inches and SD of 2 inches
Males: mean of 69 inches and SD of 3 inches
Now, let's find the probability of a randomly selected U.S. adult female shorter than 65 inches. This is asking us to find P(X < 65). Using the formula \(z = \frac{\text{observed}\mu}{\sigma}\) we find that:
\(z = \frac{6564}{2} = 0.50\)
Now, we have transformed P(X < 65) to P(Z < 0.50). From the table we see that P(Z < 0.50) = 0.6915. So, roughly there this a 69% chance that a randomly selected U.S. adult female would be shorter than 65 inches.
General Normal Distribution
I. Find the Probability Given the Range of Observation:
Example: Weights of 10year old Girls
The weights of 10yearold girls are known to be normally distributed with a mean of 70 pounds and a standard deviation of 13 pounds. Obtain the percentage of 10yearold girls with weight between 60 pounds and 90 pounds.
In other words, we want to find P(60 < X < 90), where X has a normal distribution with mean 70 and standard deviation 13.
Step 1. Sketch the normal curve for the variable.
Step 2. Shade the region of interest and mark the delimiting xvalues.
Step 3. Compute the zscores of the delimiting xvalues using the formula:
\(z=\frac{x\mu}{\sigma}\)
\(x=60,\ z=\frac{60\mu}{\sigma}=\frac{6070}{13}=0.77\)
\(x=90,\ z=\frac{90\mu}{\sigma}=\frac{9070}{13}=1.54\)
Step 4. Find the probability of Z within the delimiting zscores.
Note: The probability corresponds to the area under the standard normal curve between the two values.
\[\begin{align}P(60 < X < 90)&= P(0.77 < Z < 1.54)\\& = P(Z < 1.54)  P(Z \leq  0.77)\\& = 0.9382  0.2206 \\&= 0.7176\\ \end{align}\]
Answer: We obtain that 71.76% of 10yearold girls have weight between 60 pounds and 90 pounds.
Here is another example. Click on the link to see a walk through that helps solve the following problem:
VIEWLET: Find a value for Z, say z_{0}, such that (z_{0} < Z < z_{0}) = 0.90
Recall the empirical rule: For any data set having approximately a bellshaped distribution, we know that:
 Roughly 68% of the observations lie within one standard deviation to either side of the mean.
 Roughly 95% of the observations lie within two standard deviations to either side of the mean.
 Roughly 99.7% of the observations lie within three standard deviations to either side of the mean.
Use the normal table to validate the empirical rule. Work out your answer first, then click the graphic to compare answers.
Use the normal table to validate the empirical rule.
Answer: One standard deviation to either side of the mean implies the z score of 1 and 1.
\(P(1<Z<1)=P(Z<1)P(Z \leq 1)=0.84130.1587=0.6826\)
Similarly,
\(P(2<Z<2)=P(Z<2)P(Z \leq 2)=0.97720.0228=0.9544\)
\(P(3<Z<3)=P(Z<3)P(Z \leq 3)=0.99870.0013=0.9974\)
II. Finding the Range of Observation Given the Probability:
Example: Weights of 10year old Girls, cont'd
Find the 60th percentile of weight of 10yearold girls given that the weight is normally distributed with mean 70 pounds and standard deviation of 13 pounds.
Step 1. Sketch the normal curve for the variable.
Step 2. Shade the region of interest.
Step 3. Use the Standard Normal Cumulative Probability Table to find the zscores given the probability.
Area to the left of zscores = 0.6000.
The closest value in the table is 0.5987.
The zscore corresponding to 0.5987 is 0.25.
Thus, the 60th percentile is z = 0.25.
Step 4. Find the xvalues corresponding to the delimiting zscores found in step 3 by:
\(x = \mu + \sigma \times z\)
\(x = 70 + 13 \times 0.25 = 73.25\)
Answer: The 60th percentile of 10yearold girls' weight is 73.25 pounds.
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