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iPhone apps For the population of individuals who own an iPhone, suppose \(p=0.25\) is the proportion that has a given app. For a particular iPhone owner, let \(x=1\) if they have the app and \(x=0\) otherwise. For a random sample of 50 people who have an iPhone: a. State the population distribution (that is, the probability distribution of \(X\) for each observation). b. Find the mean of the sampling distribution of the sample proportion who have the app among the 50 people. c. Find the standard deviation of the sampling distribution of the sample proportion who have the app among the 50 people. d. Explain what the standard deviation in part \(\mathrm{c}\) describes.

Short Answer

Expert verified
a. Bernoulli distribution with p=0.25; b. Mean = 0.25; c. Standard deviation ≈ 0.0612; d. Describes spread around the population proportion.

Step by step solution

01

Population Distribution

The variable \(X\) can be described as a Bernoulli random variable because it is binary; \(x = 1\) if the individual has the app and \(x = 0\) if they do not. A Bernoulli distribution with probability \(p\) is the same as stating that \(P(X = 1) = p\) and \(P(X = 0) = 1 - p\) for a single trial. For this situation, since \(p = 0.25\), the probability distribution is \(X \sim \text{Bernoulli}(0.25)\).
02

Mean of the Sampling Distribution

The mean of the sampling distribution of the sample proportion \(\hat{p}\) for sample size \(n = 50\) can be found using the formula \(\mu_{\hat{p}} = p\). With \(p = 0.25\), the mean of the sampling distribution is \(0.25\).
03

Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of \(\hat{p}\) can be calculated using the formula \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Substituting \(p = 0.25\) and \(n = 50\) gives \(\sigma_{\hat{p}} = \sqrt{\frac{0.25(1-0.25)}{50}} = \sqrt{\frac{0.25 \times 0.75}{50}} = \sqrt{\frac{0.1875}{50}} = \sqrt{0.00375} \approx 0.0612\).
04

Interpretation of Standard Deviation

The standard deviation found in Step 3, approximately \(0.0612\), describes the expected variability or spread of the sample proportion \(\hat{p}\) around the population proportion \(p = 0.25\) when we conduct many independent random samples of 50 individuals.

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Key Concepts

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

Bernoulli Distribution
In statistics, a Bernoulli distribution is used to model a random variable that has exactly two outcomes: success or failure. This makes it perfect for scenarios where we want to know if an event occurs or not, like flipping a coin or checking if someone has an app installed on their phone. For our exercise, if an individual has the app, it is considered a "success" and is represented by the number 1. If they do not have the app, it is a "failure" and is represented by the number 0. In mathematical terms, we say that the random variable \(X\) follows a Bernoulli distribution with a probability \(p\) of success. This is denoted as \(X \sim \text{Bernoulli}(p)\). In our example, \(p = 0.25\), so \(P(X = 1) = 0.25\) and \(P(X = 0) = 0.75\).
  • Model binary outcomes
  • "Success" = 1, "Failure" = 0
  • Expressed as \(X \sim \text{Bernoulli}(p)\)
Bernoulli distributions are simple yet powerful tools for understanding binary data.
Standard Deviation
Standard deviation is a measure that describes the amount of variability or spread in a set of data. It's a crucial concept when dealing with sampling distributions because it helps us understand how much the sample proportion might deviate from the true population proportion.For a sampling distribution of a sample proportion \(\hat{p}\), the formula for standard deviation is given by:\[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\]Here, \(p\) is the probability of success, \(1-p\) is the probability of failure, and \(n\) is the sample size. In our example with \(n = 50\) and \(p = 0.25\), substituting these values gives:\[\sigma_{\hat{p}} = \sqrt{\frac{0.25 \times 0.75}{50}} = \sqrt{0.00375} \approx 0.0612\]This number, 0.0612, tells us that if we repeatedly take samples of 50 people, the proportion of those who have the app will usually be within 0.0612 of the population proportion \(p = 0.25\).
  • Measures variability or spread
  • Critical for understanding sampling accuracy
  • Helps gauge reliability of the sample proportion
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In simpler words, it tells us how likely different outcomes are.For our scenario, each individual outcome (i.e., having or not having the app) adheres to a probability distribution. As mentioned earlier, the Bernoulli distribution with \(p = 0.25\) provides a framework to understand these outcomes for each person sampled.But when we discuss the entire sample, a different type of probability distribution comes into play. We're interested in how the proportion of people who have the app behaves as we draw sample after sample. This forms a sampling distribution of the sample proportion.
  • Describes how probabilities are spread across potential outcomes
  • Essential for understanding and predicting real-world observations
  • Allows comparison between observed data and theoretical models
Probability distributions serve as foundational tools in statistics for analyzing data and making predictions.

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Most popular questions from this chapter

Using control charts to assess quality In many industrial production processes, measurements are made periodically on critical characteristics to ensure that the process is operating properly. Observations vary from item to item being produced, perhaps reflecting variability in material used in the process and/or variability in the way a person operates machinery used in the process. There is usually a target mean for the observations, which represents the long-run mean of the observations when the process is operating properly. There is also a target standard deviation for how observations should vary around that mean if the process is operating properly. A control chart is a method for plotting data collected over time to monitor whether the process is operating within the limits of expected variation. A control chart that plots sample means over time is called an \(\overline{\boldsymbol{x}}\) -chart. As shown in the following, the horizontal axis is the time scale and the vertical axis shows possible sample mean values. The horizontal line in the middle of the chart shows the target for the true mean. The upper and lower lines are called the upper control limit and lower control limit, denoted by UCL and \(\mathbf{L C L .}\) These are usually drawn 3 standard deviations above and below the target value. The region between the LCL and UCL contains the values that theory predicts for the sample mean when the process is in control. When a sample mean falls above the UCL or below the LCL, it indicates that something may have gone wrong in the production process. a. Walter Shewhart invented this method in 1924 at Bell Labs. He suggested using 3 standard deviations in setting the UCL and LCL to achieve a balance between having the chart fail to diagnose a problem and having it indicate a problem when none actually existed. If the process is working properly ("in statistical control") and if \(n\) is large enough that \(\bar{x}\) has approximately a normal distribution, what is the probability that it indicates a problem when none exists? (That is, what's the probability a sample mean will be at least 3 standard deviations from the target, when that target is the true mean?) b. What would the probability of falsely indicating a problem be if we used 2 standard deviations instead for the UCL and LCL? c. When about nine sample means in a row fall on the same side of the target for the mean in a control chart, this is an indication of a potential problem, such as a shift up or a shift down in the true mean relative to the target value. If the process is actually in control and has a normal distribution around that mean, what is the probability that the next nine sample means in a row would (i) all fall above the mean and (ii) all fall above or all fall below the mean? (Hint: Use the binomial distribution, treating the successive observations as independent.)

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