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Chapter 6: Problem 21

Scores on a common final exam in a large enrollment, multiple-section freshman course are normally distributed with mean 72.7 and standard deviation \(13.1 .\) a. Find the probability that the score \(X\) on a randomly selected exam paper is between 70 and 80 . b. Find the probability that the mean score \(x-\) of 38 randomly selected exam papers is between 70 and 80 .

Short Answer

Expert verified
(a) Probability: 0.292. (b) Probability: 0.8977.

Step by step solution

01

Understanding the Problem

We are given a normally distributed set of exam scores with a mean \(\mu = 72.7\) and a standard deviation \(\sigma = 13.1\). We need to find the probability for two different scenarios: (a) Finding the probability of a single score being between 70 and 80, and (b) Finding the probability of the mean score of a sample of 38 examinations being between 70 and 80.
02

Calculating Z-Scores for Part (a)

To find the probability for part (a), convert the scores of 70 and 80 into Z-scores using the formula \( Z = \frac{X - \mu}{\sigma} \). For \( X = 70 \), \( Z = \frac{70 - 72.7}{13.1} \approx -0.206\). For \( X = 80 \), \( Z = \frac{80 - 72.7}{13.1} \approx 0.557\).
03

Finding Probability for Part (a)

Use a standard normal distribution table or calculator to find the probabilities for the Z-scores. The probability \( P(Z < 0.557) \approx 0.711\) and \( P(Z < -0.206) \approx 0.419\). Thus, the probability that \(X\) is between 70 and 80 is \( P(-0.206 < Z < 0.557) = 0.711 - 0.419 = 0.292 \).
04

Calculating Standard Error for Part (b)

For part (b), calculate the standard error of the mean using \( SE = \frac{\sigma}{\sqrt{n}} \). Here, \( n = 38 \), so \( SE = \frac{13.1}{\sqrt{38}} \approx 2.125 \).
05

Calculating Z-Scores for Part (b)

Convert the sample mean scores 70 and 80 into Z-scores using the new standard error: \( Z_{70} = \frac{70 - 72.7}{2.125} \approx -1.271 \) and \( Z_{80} = \frac{80 - 72.7}{2.125} \approx 3.443 \).
06

Finding Probability for Part (b)

Determine the probabilities using the Z-scores: \( P(Z < 3.443) \approx 0.9997 \) and \( P(Z < -1.271) \approx 0.102 \). Thus, the probability that \( \bar{X} \) is between 70 and 80 is \( P(-1.271 < Z < 3.443) = 0.9997 - 0.102 = 0.8977 \).

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Key Concepts

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

z-scores
Z-scores are a way to measure how far away a particular score is from the mean, expressed in terms of standard deviations. When dealing with a normal distribution, Z-scores help us figure out where a particular value stands in relation to the rest of the data.
To calculate a Z-score, you use the formula \( Z = \frac{X - \mu}{\sigma} \) where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. By converting values into Z-scores, we can use standard normal distribution tables to find probabilities associated with those scores more easily.
  • Example: If the mean score of an exam is 72.7 with a standard deviation of 13.1, and you want to know how a score of 70 compares, you calculate the Z-score as \( Z = \frac{70 - 72.7}{13.1} \approx -0.206 \).
  • Positive Z-scores indicate values above the mean, while negative Z-scores indicate values below the mean. A Z-score of zero represents a value exactly at the mean.
standard deviation
Standard deviation is a measure of how spread out the values are in a dataset. It gives us an idea about the variability or dispersion of the data points. A low standard deviation means that the data points are close to the mean, while a high standard deviation indicates more spread out data points.
The standard deviation is crucial in calculating Z-scores and involves every data point in the dataset. For a normally distributed dataset, approximately 68% of the data will fall within one standard deviation of the mean. This coverage increases to about 95% within two standard deviations, and about 99.7% within three standard deviations.
  • Calculating the Standard Deviation: The formula used is \( \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \), where \( N \) is the number of data points, \( X_i \) is each value, and \( \mu \) is the mean.
  • Understanding standard deviation is essential for evaluating how likely a specific outcome is if the data follows a normal distribution.
standard error of the mean
The standard error of the mean (SEM) is a statistic that tells us how much the sample mean would vary from the actual population mean if you took multiple samples. SEM essentially measures the accuracy of the sample mean as an estimate of the population mean.
To calculate the SEM, use the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the population and \( n \) is the sample size. As the sample size increases, the standard error decreases, meaning the sample mean becomes more accurate.
  • Importance of SEM: It is crucial when conducting surveys and experiments because it helps in understanding how much uncertainty there is in the sample mean.
  • Example: If the standard deviation of exam scores is 13.1 and you have taken a sample of 38 exams, the SEM would be \( \frac{13.1}{\sqrt{38}} \approx 2.125 \).
Understanding the standard error helps decide how confident one can be in making inferences about the overall population based on a sample.

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Most popular questions from this chapter

A normally distributed population has mean 57.7 and standard deviation \(12.1 .\) a. Find the probability that a single randomly selected element \(X\) of the population is less than \(45 .\) b. Find the mean and standard deviation of \(x-\) for samples of size 16 . c. Find the probability that the mean of a sample of size 16 drawn from this population is less than 45

The proportion of a population with a characteristic of interest is \(p=0.37\). Find the mean and standard deviation of the sample proportion \(\hat{\boldsymbol{P}}\) obtained from random samples of size 125 .

A normally distributed population has mean 1,214 and standard deviation 122 . a. Find the probability that a single randomly selected element \(X\) of the population is between 1,100 and 1,300 b. Find the mean and standard deviation of \(x-\) for samples of size \(25 .\) c. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300

Suppose the mean length of time that a caller is placed on hold when telephoning a customer service center is 23.8 seconds, with standard deviation 4.6 seconds. Find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean.

The proportion of a population with a characteristic of interest is \(p=0.76\). Find the mean and standard deviation of the sample proportion \(\hat{p}\) obtained from random samples of size 1,200 .
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