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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

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. Customer Service A random sample of weekly work logs at an automobile repair station was obtained, and the average number of customers per day was recorded. $$\begin{array}{lllll} \hline 26 & 24 & 22 & 25 & 23 \\ \hline 24 & 25 & 23 & 25 & 22 \\ \hline 21 & 26 & 24 & 23 & 24 \\ \hline 25 & 24 & 25 & 24 & 25 \\ \hline 26 & 21 & 22 & 24 & 24 \\ \hline \end{array}$$

Assume that the random variable \(X\) is normally distributed, with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. \(P(56 \leq X<66)\)

The lengths of human pregnancies are approximately normally distributed, with mean \(\mu=266\) days and standard deviation \(\sigma=16\) days. (a) What proportion of pregnancies lasts more than 270 days? (b) What proportion of pregnancies lasts less than 250 days? (c) What proportion of pregnancies lasts between 240 and 280 days? (d) What is the probability that a randomly selected pregnancy lasts more than 280 days? (e) What is the probability that a randomly selected pregnancy lasts no more than 245 days? (f) A "very preterm" baby is one whose gestation period is less than 224 days. Are very preterm babies unusual?

General Electric manufactures a decorative Crystal Clear 60 -watt light bulb that it advertises will last 1500 hours. Suppose that the lifetimes of the light bulbs are approximately normally distributed, with a mean of 1550 hours and a standard deviation of 57 hours. (a) What proportion of the light bulbs will last less than the advertised time? (b) What proportion of the light bulbs will last more than 1650 hours? (c) What is the probability that a randomly selected GE Crystal Clear 60 -watt light bulb will last between 1625 and 1725 hours? (d) What is the probability that a randomly selected GE Crystal Clear 60 -watt light bulb will last longer than 1400 hours?

Steel rods are manufactured with a mean length of 25 centimeters \((\mathrm{cm}) .\) Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of \(0.07 \mathrm{~cm} .\) (a) What proportion of rods has a length less than \(24.9 \mathrm{~cm} ?\) (b) Any rods that are shorter than \(24.85 \mathrm{~cm}\) or longer than \(25.15 \mathrm{~cm}\) are discarded. What proportion of rods will be discarded? (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between \(24.9 \mathrm{~cm}\) and \(25.1 \mathrm{~cm} ?\)
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