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According to the Pew Research Center, \(53 \%\) of millennials (those born between 1981 and 1997 ) reported using a library or bookmobile within the last year. Suppose that a random sample of 200 millennials is taken. a. Complete this sentence: We would expect _____ of the sample to have used a library or bookmobile within the last year, give or take ______. b. Would it be surprising to find that 190 of the sample have used a library or bookmobile within the last year? Why or why not?

Short Answer

Expert verified
Expected: 106 millennials, give or take 7. Yes, it would be surprising to find 190 millennials have used a library, as it is a lot greater than our expected and beyond the standard deviation of 7.

Step by step solution

01

Calculate the Expected Value

The expected value, E, is calculated by multiplying the size of the sample, n, by the probability, p, of the event. In this case, n = 200 and p = 0.53. Hence, \(E = np = 200 * 0.53 = 106\). So, we would expect about 106 millennials to have used a library or bookmobile within the last year.
02

Calculate the Standard Deviation

The standard deviation, \( \sigma \), is calculated using the formula \( \sigma = \sqrt{np(1-p)} \), where n is the sample size and p is the probability of the event. Hence, \( \sigma = \sqrt{200*0.53*(1-0.53)} = 7.14 \). Therefore, we would expect about 106 of the sample to have used a library or bookmobile within the last year, give or take 7.
03

Evaluate the Observed Value

With a finding that 190 of the sample have used a library, we can compare this with our expected value and standard deviation. This observed value is far beyond 'give or take 7' from the expected 106. It is approximately 11.75 standard deviations away from the mean ( \((190-106)/7.14 = 11.75\). This deviation is quite large and, under normal circumstances, only about 5% of sample means are expected to fall more than 2 standard deviations from the mean. Therefore, it would indeed be surprising to find that 190 of the sample have used a library or bookmobile within the last year.

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Key Concepts

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

Expected Value
The expected value is a fundamental concept in probability and statistics that provides a measure of the center, or average outcome, of a random variable. It's essentially what you would anticipate as the average outcome if you could repeat an experiment an infinite number of times.

In the context of a binomial distribution, the expected value is calculated by multiplying the number of trials (the sample size) by the probability of success in a single trial. In our exercise, with a sample size of 200 millennials and a success probability of 53% using a library, the expected value, denoted as E, is calculated as follows:
\[ E = n \times p = 200 \times 0.53 = 106 \]
This indicates that, in a typical group of 200 millennials, one would expect around 106 individuals to have used a library or bookmobile within the last year.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the situation of a binomial distribution, it helps us understand how much the number of successes can differ from the expected value.

To calculate the standard deviation (\( \(\sigma\) \)) for a binomial distribution, you use the square root of the product of the sample size (n), the probability of success (p), and the probability of failure (1-p). Following our exercise:
\[ \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{200 \times 0.53 \times (1 - 0.53)} = 7.14 \]
This value gives us a sense of the 'give or take' around our expected value — in this case, about 7 millennials. Standard deviation is essential because it allows us to set expectations about how much variation we might see in the real world.
Sample Size
Sample size is the number of observations in a sample and is directly proportional to the accuracy of the estimates we can make about a population. In statistical terms, the larger the sample size, the smaller the standard deviation, which implies that our estimates are more likely to be near the true population value.

When considering sample size in the context of binomial distribution, it's important to remember that it affects both the expected value and the standard deviation. For instance, changing the sample size in our exercise would alter both the expected number of millennials using a library and the expected variation (standard deviation) around that number. As such, choosing an appropriate sample size is critical for achieving reliable statistical results.
Probability
Probability is a way of quantifying the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 represents certainty. In the context of a binomial distribution, the probability pertains to the success of a single trial.

For our exercise, the probability that a millennial has used a library in the past year is 53%, or 0.53 when expressed as a decimal. This probability is key to determining the expected value and standard deviation. It defines the underlying likelihood, which is then scaled up by the sample size to predict outcomes for larger groups, as well as calculating the range of possible outcomes around the expected value.

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Most popular questions from this chapter

According to the 2017 SAT Suite of Assessments Annual Report, the average SAT math score for students in Illinois was \(556 .\) Assume the scores are Normally distributed with a standard deviation of 100 . Answer the following including an appropriately labeled and shaded Normal curve for each question. a. What percentage of Illinois Math SAT takers scored 600 or more? b. What percentage of Illinois Math SAT takers scored between 600 and 650 ? c. Suppose students who scored in the top \(5 \%\) of test takers in the state were eligible for a special scholarship program. What SAT math score would qualify students for this scholarship program?

Directions Determine whether each of the following variables would best be modeled as continuous or discrete. a. The number of cars passing through an intersection in one hour b. The weight of a person

For each situation, identify the sample size \(n\), the probability of a success \(p\), and the number of success \(x\). When asked for the probability, state the answer in the form \(b(n, p, x)\). There is no need to give the numerical value of the probability. Assume the conditions for a binomial experiment are satisfied. A 2017 Gallup poll found that \(53 \%\) of college students were very confident that their major will lead to a good job. a. If 20 college students are chosen at random, what's the probability that 12 of them were very confident their major would lead to a good job? b. If 20 college students are chosen at random, what's the probability that 10 of them are not confident that their major would lead to a good job? a. If 20 college students are chosen at random, what's the probability that 12 of them were very confident their major would lead to a good job? b. If 20 college students are chosen at random, what's the probability that 10 of them are not confident that their major would lead to a good job?

Babies in the United States have a mean birth length of \(20.5\) inches with a standard deviation of \(0.90\) inch. The shape of the distribution of birth lengths is approximately Normal. a. Find the birth length at the \(2.5\) th percentile. b, Find the birth length at the \(97.5\) th percentile. c. Find the \(z\) -score for the length at the \(2.5\) th percentile. d. Find the \(z\) -score for the length at the \(97.5\) th percentile.

Babies in the United States have a mean birth length of \(20.5\) inches with a standard deviation of \(0.90\) inch. The shape of the distribution of birth lengths is approximately Normal. a. How long is a baby born at the 20 th percentile? b. How long is a baby born at the 50 th percentile? c. How does your answer to part b compare to the mean birth length? Why should you have expected this?

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