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ACT scores The composite scores of individual students on the ACT college entrance examination in 2009 followed a Normal distribution with mean 21.1 and standard deviation 5.1 (a) What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher? Show your work. (b) Now take an SRS of 50 students who took the test. What is the probability that the mean score \(\overline{x}\) of these students is 23 or higher? Show your work.

Short Answer

Expert verified
(a) 0.3551. (b) 0.0044.

Step by step solution

01

Understanding the distribution

The distribution of ACT scores is normal with a mean (\(\mu\)) of 21.1 and a standard deviation (\(\sigma\)) of 5.1. We will use this information to find the probability of a score being 23 or higher.
02

Calculate Z-score for individual test

To find the probability that a score is 23 or higher, we first calculate the Z-score for 23. The formula for the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \]Substituting the given values, we have:\[ Z = \frac{23 - 21.1}{5.1} \approx 0.3725 \]
03

Find probability for individual test score

Using the standard normal distribution table, or a calculator, find the probability of a Z-score being 0.3725 or higher. The probability of Z being less than 0.3725 is approximately 0.6449. Thus, the probability of scoring 23 or higher is:\[ P(Z \geq 0.3725) = 1 - 0.6449 = 0.3551 \]
04

Mean Distribution for Sample of 50 students

For a sample of 50 students, the sampling distribution of the sample mean \(\overline{x}\) is normal with mean \(\mu = 21.1\) and standard deviation \(\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}\). Here, \(n=50\), so:\[ \sigma_{\overline{x}} = \frac{5.1}{\sqrt{50}} \approx 0.7217 \]
05

Calculate Z-score for sample mean

Calculate the Z-score for a sample mean of 23:\[ Z = \frac{23 - 21.1}{0.7217} \approx 2.6244 \]
06

Find probability for sample mean score

Using the standard normal distribution table or a calculator, find the probability of a Z-score being 2.6244 or higher. The probability of Z being less than 2.6244 is approximately 0.9956. Thus, the probability of the sample mean being 23 or higher is:\[ P(Z \geq 2.6244) = 1 - 0.9956 = 0.0044 \]

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Key Concepts

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

Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical around its mean. It is often referred to as the bell curve because of its shape. This distribution is characterized by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). The mean is the central point of the distribution, while the standard deviation describes the spread of the data.
In the context of the ACT scores, the scores are normally distributed with a mean of 21.1 and a standard deviation of 5.1. This indicates that most students will score around 21.1, with fewer students scoring higher or lower. Understanding the properties of the normal distribution is crucial as it allows us to make predictions about the data, such as estimating probabilities and assessing outcomes.
Z-score Calculation
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. You can calculate the Z-score using the formula:\[Z = \frac{X - \mu}{\sigma}\]where \( X \) represents the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For example, to find the Z-score for an ACT score of 23, you subtract the mean (21.1) from 23 and then divide by the standard deviation (5.1). This results in a Z-score of approximately 0.3725. The Z-score helps to identify how many standard deviations this score is from the mean, which in turn is useful for determining probabilities using the standard normal distribution table.
Sampling Distribution
The sampling distribution refers to the probability distribution of a sample statistic. When we take samples of a population, such as 50 students from all those who took the ACT, each sample can have different characteristics. The sampling distribution of the sample means specifically describes how those mean scores are distributed.
For normally distributed data, the sampling distribution of the sample mean tends to be normal as well, especially if the sample size is large enough (usually \( n \geq 30 \)). The mean of the sampling distribution is the same as the population mean (\( \mu \)), but the standard deviation is reduced, calculated as \( \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \).
In the case of 50 ACT scores, the standard deviation of the sampling distribution becomes approximately 0.7217, making the distribution of the sample means more concentrated around the mean.
Standard Normal Distribution Table
The standard normal distribution table (or Z-table) is a tool that shows the probability of a standard normal random variable being less than or equal to a given value. It is used to find probabilities for standard normal distributions, which have a mean of 0 and a standard deviation of 1.
When calculating probabilities for Z-scores, as seen in our ACT example, you look up the Z-score in the table to find the probability that a value is less than the given Z-score. For instance, a Z-score of 0.3725 corresponds to a cumulative probability of approximately 0.6449, meaning there is a 64.49% chance of scoring below 23. To find out the probability of scoring 23 or higher, you subtract this probability from 1.
This use of the table is fundamental in statistics for quickly finding probabilities and making decisions based on standard normal distributions.

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Most popular questions from this chapter

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