Question 1 (1 point)
Last week we looked at different distributions — such as bi-modal, uniform, skewed, and normal. This week we will focus our attention on one important distribution — the normal distribution. In particular, we will look at the standard normal curve. For every normal curve, we can transform its values to the standard normal curve using the formula Z = (X – Xbar)/S, where Z is the standard score, X is our value to be transformed, Xbar is the mean of the normal distribution, and S is the standard deviation. For all of these problems, it helps to visualize a standard normal curve, and where the value(s) of X fall, and how much of the area under the curve is to the left/right/middle of the value(s).
1. Drinking survey. One reason that Normal distributions are important is that they describe the results of phenomena that have a random element, such as how an opinion poll would vary if the poll were repeated many times, or for data on samples of average behavior, such as drinking.
A report by the National Study on Drug Use and Health finds that on average females consumed an average of 2.4 drinks per day in the past 30 days. And let’s assume that the data follow the Normal distribution with standard deviation 1.1 drinks per day. Use this fact and the 68-95-99.7 rule to answer the following two questions (in other words, you should not need to do any calculations to get the answers.)
In repeated samples, what percent of samples would give a result above 2.4 drinks per day for females (In English: if I knew the population mean was 2.4 and I randomly sampled values and took their average, what percent of my samples would produce a value above 2.4)? Picture the normal curve and the area above the mean.
Question 1 options:
| A. | 100% |
| B. | 50% |
| C. | -50% |
| D. | 25% |
1b. For the drinking data above, in large number of samples, if I have a mean of 2.4 and a standard deviation of 1.1, what range of number of drinks contains the central 95.45% of number of drinks per day? (You do not need to do calculations, just remember the Rule you’ve learned this week.)
Question 2 options:
| A. | 0.2 to 4.6 drinks |
| B. | 2.4 drinks |
| C. | 1.3 to 3.5 drinks |
| D. | 95% |
2. For the following four questions, find some proportions (or areas under the curve). You need to become familiar with standardizing survey results by transforming them into Z (and later t) scores. Then, you have to find the probability of being to the left, right or between Z scores. The sign < means all the values (or area under the curve) to the left of the Z score. The sign > means all the values to the right of the Z score. For example, in standard z-score world, the probability that Z<0 = .50, or 50%. The probability that 0<Z<1 = .3413, or 34.13%., and -1<Z<1 = .68 Try this on your own! Checkout https://www.freecodecamp.org/news/normal-distribution-explained/ for more info.
Using either Excel’s =normsdist() function, or your book, or your z-score calculator from this week’s forum or software (or drawing a picture often helps), find the proportion of observations from a standard Normal distribution (bell curve) for each of the following events, i.e., what proportion of the curve is in that area:
a.Probability that Z > 1.96 (almost 2 or more standard deviations above the mean)
Question 3 options:
| A. | -.233 |
| B. | .05 |
| C. | .025 |
| D. | .975 |
2b. Probability that Z < 1.96 (or that the value is less than a 2 standard deviation – think about whether most values are more or less than 2 standard deviations from the mean)
Question 4 options:
| A. | .975 |
| B. | .025 |
| C. | .5 |
| D. | .33 |
2c. Probability that Z > 1.28 (or Z< -1.28)
Question 5 options:
| A. | 0.05 |
| B. | 0.5 |
| C. | 0.90 |
| D. | 0.1003 |
2d. Probability that -1.96 > Z > 1.96 (this is the probability that the value is less than negative 1.96 AND greater than 1.96). This is the value in the two tails.
Question 6 options:
| A. | 0.8747 |
| B. | 0.1003 |
| C. | 0.05 |
| D. | 1.0 |
3. The Online police department was asked by the mayor’s office to estimate the cost of crime to citizens of Online. The police began their study with the crime of identity theft, taking a random sample of files (there is too much crime to calculate the statistics for all the crimes committed). They found the average dollar loss in an identity theft was $6000, with a standard deviation of $2000, and that the dollar loss was normally distributed. (Use this information for the following 4 questions. Again, calculate a z-score, and then use a table or app to figure out the percentage. Some apps will do it all for you if you enter the mean and standard deviation.) In this sample:
- What proportion of identity thefts had dollar losses above $10000?
Question 7 options:
| A. | 0.60 |
| B. | 0.2743 |
| C. | .0228 |
| D. | 5000 |
3b. If you have a normal distribution with mean=6000 and standard deviation =2000, what percentage of identity thefts have dollar losses between $2000 and $4000?
Question 8 options:
| A. | 0.9394 |
| B. | 0.1359 |
| C. | 0.0991 |
| D. | 0.2528 |
c. If you have a normal distribution with mean=6000 and standard deviation =2000, what is the probability that any one identity theft had a dollar loss above $4000?
Question 9 options:
| A. | 0.6808 |
| B. | 0.15865 |
| C. | .841345 |
| D. | .025 |
Question 10 (1 point)
Why all these questions about normal distributions this week (with Z scores and probabilities)? Where are we going with these lessons?
How in the heck does this relate to public administration or your world? Any concrete examples?
Question 10 options (SHORT ANSWERS)
4. In an “Excite Poll” October 14, 2002, a self-response poll of internet users, the question was:
It is now possible for school students to log on to Internet sites and download homework. Everything from book reports to doctoral dissertations can be downloaded free or for a fee. Do you believe giving a student who is caught plagiarizing an “F” for their assignment is the right punishment?
In all, 14,793 people clicked “Yes,” 1778 clicked “No, it is too harsh,” 2566 clicked “No, it is not harsh enough,” and 988 clicked “don’t know” or “don’t care.”
a) What is the sample size for this poll?
Question 11 options:
| A. | 1778 |
| B. | 2566 |
| C. | 4344 |
| D. | 20125 |
| E. | 14793 |
Question 12 (0.5 points)
4b.
The “Excite Poll” internet write in poll described above has a much larger sample than standard sample surveys. Can we trust the result to give good information about any clearly defined population. Why or why not? Would a random sample of 50 provide better or worse data?
Question 12 options (SHORT ANSWER)
GROUP #1
Building on what you chose as the most important factors affecting retention from last week (that were in the retention dataset), produce some basic descriptive statistics or charts using the 1000 or so cases associated with your group number — e.g., frequency tables, pie charts, or histograms/bar charts for your variables of interest.
For example, if you chose GPA, what is it’s frequency distribution? What is the mean/median GPA?
Use Excel formulas to calculate the mean, median, and mode.
a. Compare the mean, median and mode for your variable, and explain why they differ (or why not).
b. Which number — mean, median, or mode — best describes the center of each distribution?
2. Calculate the range and standard deviation.
3. Create a graph or histogram for one of your variables. (You may first have to group values into categories, or bins. For example, you may group GPA as 0-1.29, 1.3-1.69, 1.7-1.99,2-2.29, and so on.
a. Describe the shape of the distribution (uniform, normal, skewed, bi-modal).
b. What does this graph tell you about how scores are bunched?
Each student may each choose to do one or more variable of interest. Upload an Excel file here. Also, next week you’ll post your answer in the Group Forum topic for Week 4.
As a group, you’ll put together your best summary data. [Hint: it’s not a bad idea to include at least two or three interval level data variables, as a group.]
