| Probability and Sampling Distributions |
| 1.Thinking about probability statements. Probability is measure of how likely an event is to occur. Match one of probabilities that follow with each statement of likelihood given (The probability is usually a more exact measure of likelihood than is the verbal statement.) | | | | | | | | | | | Answer |
| 0 0.01 0.3 0.6 0.99 1 |
| (a) This event is impossible. It can never occur. |
| (b) This event is certain. It will occur on every trial. |
| (c) This event is very unlikely, but it will occur once in a while in a long sequence of trials. |
| (d) This event will occur more often that not. |
| 2. Spill or Spell? Spell-checking software catches "nonword errors" that result in a string of letters that is not a word, as when "the" is typed as "the." When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution: |
| | Value of X | | 0 | 1 | 2 | 3 | 4 |
| | Probability | | 0.1 | 0.2 | 0.3 | 0.3 | 0.1 |
| (a) Check that this distribution satisfies the two requirements for a legitimate assignment of probabilities to individual outcomes. |
| (b) Write the event "at least one nonword error" in term of X (for example, P(X >3)). What is the probability of this event? |
| (c) Describe the event X ≤ 2 in words. What is its probability? |
| 3. Discrete or continuous? For each exercise listed below, decide whether the random variable described is discrete or continuous and explains the sample space. |
| (a) Choose a student in your class at random. Ask how much time that student spent studying during the past 24 hours. |
| (b) In a test of a new package design, you drop a carton of a dozen eggs from a height of 1 foot and count the number of broken eggs. |
| (c) A nutrition researcher feeds a new diet to a young male white rat. The response variable is the weight (in grams) that the rat gains in 8 weeks. |
| 4. Tossing Coins |
| (a) The distribution of the count X of heads in a single coin toss will be as follows. Find the mean number of heads and the variance for a single coin toss. |
| | Number of Heads (Xi) | | | 0 | | 1 | | | | mean: |
| | Probability (Pi) | | | 0.5 | | 0.5 | | | | variance: |
| (b) The distribution of the count X of heads in four tosses of a balanced coin was as follows but some missing probabilities. Fill in the blanks and then find the mean number of heads and the variance for the distribution with assumption that the tosses are independent of each other. |
| | Number of Heads (Xi) | | | 0 | 1 | 2 | 3 | 4 | | mean: |
| | Probability (Pi) | | | 0.0625 | | | | 0.0625 | | variance: |
| (c) Show that the two results of the means (i.e. single toss and four tosses) are related by the addition rule for means. |
| (d) Show that the two results of the variances (i.e. single toss and four tosses) are related by the addition rule for variances (note: It was assumed that the tosses are independent of each other). |
| 5. Generating a sampling distribution. Let's illustrate the idea of a sampling distribution in the case of a very small sample from a very small population. The population is the sizes of 10 medium-sized business where size is measured in terms of the number of employees. For convenience, the 10 companies have been labeled with the integers 0 to 9. |
| Company | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Size | 82 | 62 | 80 | 58 | 72 | 73 | 65 | 66 | 74 | 62 |
| The parameter of interest is the mean size μ in this population. The sample is an SRS of size = 4 drawn from the population. Because the companies are labeled 0 to 9, a single random digits from Table B chooses one company for the sample. |
| (a) Find the mean of the 10 sizes in the population. This is the population mean, μ. |
| (b) Use Table B Line 107, 108 and 109 – 82739 57890 20807 47511 81676 55300 94383 14893
60940 72024 17868 24943 61790 90656 87964 18883
36009 19365 15412 39638 85453 46816 83485 41979
- to draw an SRS of size 4 from this population. Write the four sizes in your samples and calculate the mean of your sample. This statistic is an estimate of μ.
(Note: The company is labeled from 0 to 9, you can use a single random digit) |
| sample: | | | | | | | | | | | sample mean: |
| (c) Repeat this process 10 times using different parts of Table B. What is the mean of these 10 sample means? Is it close to the population mean, μ? This process is the construction of sampling distribution (even if we need a histogram).) |
| sample #1: | | | | | | | | | | | 1st sample mean: |
| sample #2: | | | | | | | | | | | 2nd sample mean: |
| sample #3: | | | | | | | | | | | 3rd sample mean: |
| sample #4: | | | | | | | | | | | 4th sample mean: |
| sample #5: | | | | | | | | | | | 5th sample mean: |
| sample #6: | | | | | | | | | | | 6th sample mean: |
| sample #7: | | | | | | | | | | | 7th sample mean: |
| sample #8: | | | | | | | | | | | 8th sample mean: |
| sample #9: | | | | | | | | | | | 9th sample mean: |
| sample #10: | | | | | | | | | | | 10th sample mean: |
| | | | | | | | | | | | Grand mean
(Mean of sample means) |
| Is the mean of sample means (grand mean) close to the population mean? |
| 6. The Cost of Internet Access. The amount that households pay service providers for access to the Internet varies quite a bit, the mean monthly fee is $28 and the standard deviation is $10. The distribution is NOT Normal: many households pay about $10 for limited dial-up access or about $25 for unlimited dial-up access, but many pay more for broadband connections. A sample survey asks an SRS of 400 households with Internet access how much they pay. |
| (a) Let X bar be the mean monthly fee of Internet access. What is the approximate distribution of X bar according to the Central Limit Theorem? In other words, what is the sampling distribution of X bar? |
| (b) What is the probability that the average fee paid by the sample households exceeds $29? |
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