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Chapter 7: Problem 56

The ACT and SAT are two college entrance exams. The composite score on the ACT is approximately normally distributed with mean 21.1 and standard deviation 5.1 . The composite score on the SAT is approximately normally distributed with mean 1026 and standard deviation \(210 .\) Suppose you scored 26 on the \(\mathrm{ACT}\) and 1240 on the SAT. Which exam did you score better on? Justify your reasoning using the normal model.

Short Answer

Expert verified
The SAT score is better (Z=1.02) than the ACT score (Z=0.96).

Step by step solution

01

- Understand the Problem

We need to determine on which test the provided scores perform better compared to their respective means and standard deviations.
02

- Convert ACT Score to Z-Score

Calculate the Z-score for the ACT. The Z-score formula is \[Z = \frac{X - \mu}{\sigma}\]where \(X\) is the score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For the ACT, \(X = 26\), \(\mu = 21.1\), and \(\sigma = 5.1\):\[Z_{ACT} = \frac{26 - 21.1}{5.1} = 0.96\]
03

- Convert SAT Score to Z-Score

Calculate the Z-score for the SAT. Using the same Z-score formula, for the SAT, \(X = 1240\), \(\mu = 1026\), and \(\sigma = 210\):\[Z_{SAT} = \frac{1240 - 1026}{210} = 1.02\]
04

- Compare the Z-Scores

Compare the Z-scores obtained for both the ACT and SAT. A higher Z-score indicates a better performance relative to the distribution of scores.

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Key Concepts

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

normal distribution
In statistics, a normal distribution is a type of continuous probability distribution for a real-valued random variable. When graphed, it forms a bell-shaped curve that is symmetric about the mean. These are key characteristics of a normal distribution:
  • Symmetric around the mean.
  • Mean, median, and mode are all equal.
  • Follows the 68-95-99.7 rule:
This means that approximately 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In the context of the ACT and SAT scores, both are assumed to be normally distributed, allowing us to apply statistical methods to compare scores. For the ACT, the mean score is 21.1 with a standard deviation of 5.1, while the SAT has a mean score of 1026 and a standard deviation of 210. This normality assumption is crucial for calculating Z-scores.
Z-score calculation
A Z-score, or standard score, indicates how many standard deviations an element is from the mean. It is a way to compare scores from different distributions by standardizing them. The formula for calculating a Z-score is:

\[Z = \frac{X - \mu}{\sigma}\]

where:
  • \(X\) is the value you are standardizing,
  • \(\mu\) is the mean of the distribution,
  • \(\sigma\) is the standard deviation of the distribution.
Let's apply this to the scores:
  • For the ACT: \[Z_{ACT} = \frac{26 - 21.1}{5.1} = 0.96\]
  • For the SAT: \[Z_{SAT} = \frac{1240 - 1026}{210} = 1.02\]
A higher Z-score implies the score is farther from the mean in the positive direction, indicating better performance compared to peers.
statistical comparison
Once we calculate the Z-scores for both tests, we can use them to make a statistical comparison about the performance relative to the respective test populations. Z-scores help us understand whether a given score is above or below the average and by how much compared to the spread of the data (standard deviation).
In this problem, the Z-scores are:
  • ACT: 0.96
  • SAT: 1.02

Since 1.02 > 0.96, the SAT score of 1240 is better in terms of deviation from the mean than the ACT score of 26.
Therefore, you performed relatively better on the SAT compared to the ACT. Converting raw scores to Z-scores has allowed us to make a fair comparison between the two different distributions.

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Most popular questions from this chapter

Monthly charges for cell phone plans in the United States are normally distributed with mean \(\mu=\$ 62\) and standard deviation \(\sigma=\$ 18 .\) (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of plans that charge less than \(\$ 44\) (c) Suppose the area under the normal curve to the left of \(x=\$ 44\) is 0.1587 . Provide two interpretations of this result.

The mean incubation time of fertilized chicken eggs kept at \(100.5^{\circ} \mathrm{F}\) in a still-air incubator is 21 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day. Source: University of Illinois Extension (a) Draw a normal model that describes egg incubation times of fertilized chicken eggs. (b) Find and interpret the probability that a randomly selected fertilized chicken egg hatches in less than 20 days. (c) Find and interpret the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch. (d) Find and interpret the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days. (e) Would it be unusual for an egg to hatch in less than 18 days? Why?

The birth weights of full-term babies are normally distributed with mean \(\mu=3400\) grams and \(\sigma=505\) grams. Source: Based on data obtained from the National Vital Statistics Report, Vol. \(48,\) No. 3 (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of full-term babies who weigh more than 4410 grams. (c) Suppose the area under the normal curve to the right of \(x=4410\) is \(0.0228 .\) Provide two interpretations of this result.

The number of chocolate chips in an 18 -ounce bag of Chips Ahoy! chocolate chip cookies is approximately normally distributed, with a mean of 1262 chips and a standard deviation of 118 chips, according to a study by cadets of the U.S. Air Force Academy. Source: Brad Warner and Jim Rutledge, Chance \(12(1): 10-14,1999\) (a) Determine the 30 th percentile for the number of chocolate chips in an 18 -ounce bag of Chips Ahoy! cookies. (b) Determine the number of chocolate chips in a bag of Chips Ahoy! that make up the middle \(99 \%\) of bags. (c) What is the interquartile range of the number of chips in Chips Ahoy! cookies?

Find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the left of (a) \(z=-3.49\) (b) \(z=-1.99\) (c) \(z=0.92\) (d) \(z=2.90\)
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