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Based on an LG smartphone survey, assume that \(51 \%\) of adults with smartphones use them in theaters. In a separate survey of 250 adults with smartphones, it is found that 109 use them in theaters. a. If the \(51 \%\) rate is correct, find the probability of getting 109 or fewer smartphone owners who use them in theaters. b. Is the result of 109 significantly low?

Short Answer

Expert verified
a. The probability is approximately 0.0096. b. Yes, 109 is significantly low.

Step by step solution

01

Define the Parameters

Identify the key variables of the problem. Let the population proportion be denoted as \(p\). Given \(p = 0.51\), and the sample size \(n = 250.\)
02

Determine Mean and Standard Deviation

Calculate the mean \(\mu\) and standard deviation \(\sigma\) for the binomial distribution. Use the formulas: \[ \mu = np \] and \[ \sigma = \sqrt{np(1-p)} \] Substituting the values: \[ \mu = 250 \times 0.51 = 127.5 \] \[ \sigma = \sqrt{250 \times 0.51 \times (1 - 0.51)} \approx 7.91 \]
03

Convert to a Standard Normal Distribution

Transform the binomial variable to a standard normal variable to use the Z-table. The formula is: \[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is the number of successes. Substituting \(X = 109\): \[ Z = \frac{109 - 127.5}{7.91} \approx -2.34 \]
04

Find the Corresponding Probability

Use standard normal distribution tables or a calculator to find the probability corresponding to \(Z = -2.34\). \[ P(Z < -2.34) \approx 0.0096 \]
05

Interpret the Result

Compare the probability obtained to a significance level (commonly 0.05 or 0.01). Since \(0.0096 < 0.05\) and \(0.0096 < 0.01\), the occurrence of 109 or fewer users out of 250 is very unlikely if \(p = 0.51\).

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Key Concepts

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

standard normal distribution
The standard normal distribution, often referred to simply as the Z-distribution, is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. This distribution is used extensively in statistics because it allows us to calculate probabilities and percentiles for different values. Every normal distribution can be transformed into the standard normal distribution by using the Z-score formula: dist. formula: \[ Z = \frac{X - \text{\textmu}}{\text{\textsigma}} \]where:
- \(X\) represents the value of the random variable
- \( \text{\textmu} \) is the mean of the distribution
- \( \text{\textsigma} \) is the standard deviation of the distribution.
In our exercise with the smartphone question, we converted our binomial distribution to a normal distribution and then further standardized it using this formula. This gives us a Z-score, which tells us how many standard deviations away our observed value is from the mean.
probability calculation
Probability calculation is at the heart of statistical inference and decision-making. We calculate the probability of certain outcomes to understand how likely they are to occur. In this exercise, after determining the Z-score (approximately -2.34), we look up the probability associated with this value using a Z-table or a calculator.
The standard normal distribution table (or Z-table) provides the probability that a standard normal random variable will be less than a given value (in our case, -2.34). For instance, a Z-score of -2.34 corresponds to a probability of approximately 0.0096. This means there is a 0.96% chance that 109 or fewer smartphone users out of 250 will use their smartphones in theaters if the true usage proportion is 51%.
Remember:
  • Standardizing makes it easier to compare probabilities across different normal distributions.
  • Z-tables convert Z-scores to probabilities, which helps us make informed decisions based on statistical evidence.
population proportion
The population proportion, denoted as \( p \), represents the true fraction of the entire population that possesses a particular attribute or characteristic. In probability and statistics, we often use sample data to estimate this unknown population proportion.
In our smartphone theater example, we assume the population proportion \( p \) is 0.51 or 51%. This means that, on average, 51% of all adults with smartphones use them in theaters.
Finally, here's a quick highlight to help digest the concept:
  • Population Proportion (\( p \)): The true fraction or percentage in the population we are interested in.
  • Sample Size (\( n \)): The number of observations we have or collect from the population.
  • Significance Level: A threshold used to judge if a statistical result is unusual or not. Common levels are 5% (0.05) and 1% (0.01).
Using these concepts, we can calculate probabilities and make conclusions about how likely an observed outcome is compared to what we would expect if our population proportion estimate is correct.

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Most popular questions from this chapter

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