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Chapter 9: Problem 3

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a. Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be bigger than 80\. Shade the region under the Normal curve whose area corresponds to this probability.

Short Answer

Expert verified
The labelled curve will have 70 in the center with marks at 60, 80, 50, 90, 40, and 100. The probability that a randomly selected score is bigger than 80 is approximately 0.1587.

Step by step solution

01

Sketch the Normal curve and label the positions

Draw a symmetrical bell-shaped curve. Label the mean (70) in the center, on the \(x\)-axis. Mark the points that are one standard deviation (10) away from the mean, i.e., 60 and 80, two standard deviations away, i.e., 50 and 90, and three standard deviations away, i.e., 40 and 100.
02

Calculate the Z-score for the value 80

The Z-score is found using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Plugging in \(X = 80\), \(\mu = 70\), and \(\sigma = 10\), we get \(Z = \frac{80 - 70}{10} = 1\). This tells us that 80 is one standard deviation above the mean.
03

Find the probability

We need to find the probability that a randomly selected score is bigger than 80, i.e., the area to the right of the Z-score of 1 under the Normal curve. Using a standard Normal distribution table, look up the Z-score of 1 and subtract the value from 1 (since the table gives the area to the left of the Z-score) to get the answer. For Z = 1, the table lists the area as 0.8413. Thus, the probability would be \(1 - 0.8413 = 0.1587\). Shade this area under the curve on the sketch.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It's an essential concept in statistics, giving insight into the spread of data points around the mean. In a Normal distribution, standard deviation determines the width of the bell curve. The smaller the standard deviation, the narrower and steeper the curve. Conversely, a larger standard deviation indicates a wider and flatter curve.

In the context of the exercise, the mean score is 70, and the standard deviation is 10. This means that most students' scores lie within 10 points of the mean. Normal distributions follow a specific rule called the Empirical Rule or 68-95-99.7 rule. This rule states:
  • About 68% of data points fall within one standard deviation (from 60 to 80 in this case)
  • 95% lie within two standard deviations (50 to 90)
  • 99.7% are within three standard deviations (40 to 100)
The standard deviation helps identify how scores deviate from the mean and can hint towards outliers or unusually high or low performances.
Z-score
The Z-score, also known as the standard score, indicates how many standard deviations an element is from the mean. It is a way to standardize the data point within the distribution, allowing comparison between different datasets.

The Z-score is calculated with the formula: \[ Z = \frac{X - \mu}{\sigma} \]Where:
  • \(Z\) is the Z-score
  • \(X\) is the value of the data point
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation
In our exercise, the score of 80 yielded a Z-score of 1. This means 80 is one standard deviation above the mean score of the exam. Z-scores are valuable for understanding how extreme a particular score is in comparison to the average. They are also necessary for calculating probabilities associated with different parts of the distribution.
Probability
Probability is the measure of the likelihood that an event will occur. Under a Normal distribution, different scores have different probabilities depending on their position in the distribution. Probability calculations help determine the area under the curve for specific ranges, which represents the likelihood of obtaining those scores.

In the given problem, finding the probability that a score is greater than 80 requires the use of Z-scores and a standard Normal distribution table. After computing the Z-score for 80, which is 1, you can reference the table to find the probability corresponding to a Z-score of 1. This table provides the cumulative probability from the far left to any given Z-score.

For Z = 1, the cumulative probability is 0.8413, meaning 84.13% of students score below 80. To find the probability of scores above 80, you subtract this from 1: \[ 1 - 0.8413 = 0.1587 \]The result, 0.1587, indicates there is a 15.87% chance of randomly selecting a student who scored above 80. Understanding these probabilities can assist in gauging the commonality or rarity of specific scores in a dataset.

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Most popular questions from this chapter

Tomatoes Use the data from Exercise \(9.36\). a. Using the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5 pounds using a significance level of \(0.05\) ? Why or why not? The confidence interval is reported here: \(\mathrm{I}\) am \(95 \%\) confident the population mean is between \(4.9\) and \(5.3\) pounds. b. Now test the hypothesis that the population mean is not 5 pounds using the four step procedure. Use a significance level of \(0.05\) and number your steps.

Oranges A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed \(10.2,10.5,10.3\), and \(10.3\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. I am \(95 \%\) confident that the population mean is between ii. There is a \(95 \%\) chance that all intervals will be between iii. I am \(95 \%\) confident that the sample mean is between b. Does the interval capture 10 pounds? Is there enough evidence to reject the null hypothesis that the population mean weight is 10 pounds? Explain your answer.

GPAs (Example 11) In finding a confidence interval for a random sample of 30 students GPAs, one interval was \((2.60,3.20)\) and the other was \((2.65,3.15)\). a. One of them is a \(95 \%\) interval and one is a \(90 \%\) interval. Which is which, and how do you know? b. If we used a larger sample size \((n=120\) instead of \(n=30\) ). would the \(95 \%\) interval be wider or narrower than the one reported here?

Vegetarians' Weights The mean weight of all 20-yearold women is 128 pounds (http://www.kidsgrowth.com). A random sample of 40 vegetarian women who are 20 years old showed a sample mean of 122 pounds with a standard deviation of 15 pounds. The women's measurements were independent of each other. a. Determine whether the mean weight for 20 -year old vegetarian women is significantly less than 128 , using a significance level of \(0.05\). b. Now suppose the sample consists of 100 vegetarian women who are 20 years old, and repeat the test. c. Explain what causes the difference between the \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\).

Carrots The weights of four randomly chosen bags of horse carrots, each bag labeled 20 pounds, were \(20.5,19.8,20.8\), and \(20.0\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of horse carrots. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. \(95 \%\) of all sample means based on samples of the same size will be between and ii. I am \(95 \%\) confident that the population mean is between and iii. We are \(95 \%\) confident that the boundaries are and b. Can you reject a population mean of 20 pounds? Explain.
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