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\(1,3,5,7\), and 9 are odd and 0,2, 4,6, and 8 are even. Consider a 30 -digit line from a random number table. a. How many of the 30 digits would you expect to be odd on average? b. If you actually counted, would you get exactly the number you predicted in part a? Explain.

Short Answer

Expert verified
a. On average, one would expect 15 of the 30 digits to be odd. b. The actual count will probably not always match the expected number due to randomness in the counting process. However, over many trials, the average should approach the expected value of 15.

Step by step solution

01

Identifying Odd and Even Numbers

First, identify the set of odd and even numbers in the range 0-9. There are 5 odd numbers: {1,3,5,7,9}, and 5 even numbers: {0,2,4,6,8}.
02

Calculate the Probability of Odd Numbers

As we are considering a uniform random number table where each digit between 0 and 9 is equally likely, the probability of getting an odd number at any given position in the 30-digit line is equal to the ratio of odd numbers to total numbers. This is \( \frac{5}{10} = 0.5 \)
03

Calculate Expected Value

The expected value is the hypothetical average result of an experiment (selecting a digit in the number line) repeated many times. In this case, it's the total number of digits (30) multiplied by the probability of getting an odd number (0.5). Therefore, the Expected value = 30*0.5 = 15. We expect 15 out of 30 digits to be odd.
04

Variability around the Expectation

In part b, the question is about whether the actual count will match the expected result or not. Although the expected value is 15, this is a random process, so the actual count can vary. When repeating this experiment several times, we expect the count to vary around the value of 15. Sometimes it would be more, sometimes less, but on average, it should be approximately 15. Therefore, we wouldn't expect the number to be exactly 15 every time, due to the randomness of the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability in Statistics
Probability is a fundamental concept in statistics, representing the likelihood of a particular event occurring. It ranges from 0 (impossible) to 1 (certain). When dealing with a finite set of equally likely outcomes, such as picking a numbered ball from a bag, the probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

For example, in a random number table where every digit from 0 to 9 has an equal chance of being selected, the probability of choosing an odd number is the number of odd digits (1, 3, 5, 7, 9) divided by the total digits (0 through 9). Since there are five odd and five even numbers in this set, the probability is calculated as \( \frac{5}{10} = 0.5 \). This means if we were to pick a digit at random, there's a 50% chance it would be odd. In relation to the exercise, with a 30-digit line from a random number table, we would thus expect on average half of the digits to be odd, that is 15 out of 30.
Random Number Table
A random number table is a tool used in statistics to simulate random outcomes. It consists of a sequence of digits (usually 0-9) that are generated in a manner where each digit has an equal chance of appearing. This method ensures that there is no discernible pattern which could influence the results of an experiment based on randomness.

When working with a random number table, as in the given exercise, it is important to remember that each digit is independent of the others. That means knowing the value of one digit does not provide any information about the values of subsequent or preceding digits. The table is thus utilized to mimic the randomness you would find in real-world scenarios. If we were inspecting a 30-digit sequence from such a table to find out how many odd numbers appear, we'd employ the concept of expected value. However, while the expected value provides a mean for prediction, the 'law of large numbers' implies that the actual observed frequency will converge to the expected value only over a large number of trials.
Odd and Even Numbers
Odd and even numbers are the most basic categories of integers. An even number is any integer that can be divided by two without a remainder, such as 0, 2, 4, 6, and 8. Conversely, an odd number cannot be evenly divided by two; the set includes numbers like 1, 3, 5, 7, and 9. These properties have distinct implications in probability exercises involving randomness.

In the context of the exercise, recognizing a number as odd or even tells us its membership in one of two equally sized groups. This categorization is crucial for calculating expectations about their frequency in any random numeric sample, such as the 30-digit line from a random number table. It's important to note that while we might expect, on average, an equal distribution of odd and even numbers, any given sample could deviate from this average due to natural variability inherent to random processes.

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Most popular questions from this chapter

A double-blind study using random assignment was done of pregnant women in Denmark. Women were given fish oil or a placebo during pregnancy. Their children were followed during the first 5 years of life to see if they developed asthma. The results are summarized in the table. (Bisgaard et al., "Fish Oil-Derived Fatty Acids in Pregnancy and Wheeze and Asthma in Offspring," New England Journal of Medicine, vol. 375: 2530-2539. doi: 10.1056/NEJMoa1503734) $$ \begin{array}{|lcc|} \hline \text { Developed asthma } & \text { Fish Oil } & \text { Placebo } \\ \hline \text { Yes } & 58 & 83 \\ \hline \text { No } & 288 & 266 \\ \hline \end{array} $$ a. Calculate and compare the percentages of children who developed asthma in the fish oil group and in the placebo group. b. Check that the conditions for using a two-population confidence interval hold. c. Find the \(95 \%\) confidence interval for the difference in the proportion of children who develop asthma in the two groups. Based on your confidence interval, can we conclude that there is a difference in the population proportions?

A random sample of likely voters showed that \(49 \%\) planned to support Measure \(\mathrm{X}\). The margin of error is 3 percentage points with a \(95 \%\) confidence level. a. Using a carefully worded sentence, report the \(95 \%\) confidence interval for the percentage of voters who plan to support Measure \(X\). b. Is there evidence that Measure X will fail? c. Suppose the survey was taken on the streets of Miami and the measure was a Florida statewide measure. Explain how that would affect your conclusion.

Suppose a political consultant is hired to determine if a school bond is likely to pass in a local election. The consultant randomly samples 250 likely voters and finds that \(52 \%\) of the sample supports passing the bond. Construct a \(95 \%\) confidence interval for the proportion of voters who support the bond. Assume the conditions are met. Based on the confidence interval, should the consultant predict the bond will pass? Why or why not?

According to a 2017 Pew Research report, \(40 \%\) of millennials have a BA degree. Suppose we take a random sample of 500 millennials and find the proportion who have a BA degree. a. What value should we expect for our sample proportion? b. What is the standard error? c. Use your answers to parts a and \(\mathrm{b}\) to complete this sentence: We expect _____% to have a BA degree give or take _____%. d. Suppose we decreased the sample size from 500 to 100 . What effect would this have on the standard error? Recalculate the standard error to see if your prediction was correct.

In 2018 Gallup reported that \(52 \%\) of Americans are dissatisfied with the quality of the environment in the United States. This was based on a \(95 \%\) confidence interval with a margin of error of 4 percentage points. Assume the conditions for constructing the confidence interval are met. a. Report and interpret the confidence interval for the population proportion that are dissatisfied with the quality of the environment in the United States in 2018 . b. If the sample size were larger and the sample proportion stayed the same, would the resulting interval be wider or narrower than the one obtained in part a? c. If the confidence level were \(90 \%\) rather than \(95 \%\) and the sample proportion stayed the same, would the interval be wider or narrower than the one obtained in part a? d. In 2018 the population of the United States was roughly 327 million. If the population had been half that size, would this have changed any of the confidence intervals constructed in this problem? In other words, if the conditions for constructing a confidence interval are met, does the population size have any effect on the width of the interval?

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