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Chapter 8: Problem 25

LSAT test scores are normally distributed with a mean of 500 and a standard deviation of 100 . Find the probability that a randomly chosen test-taker will score between 300 and 550 .

Short Answer

Expert verified
The probability that a randomly chosen test-taker will score between 300 and 550 on the LSAT is approximately 0.6687, or 66.87%. This is found by calculating the z-scores for the scores and using a standard normal table to find the area under the standard normal curve between the z-scores.

Step by step solution

01

Identify the given values

In this problem, we are given the following: Mean (渭) = 500 Standard deviation (蟽) = 100 Lower score (a) = 300 Upper score (b) = 550
02

Calculate the z-scores for the given scores

To calculate the z-scores for the lower and upper scores, use the z-score formula: \(z = \frac{x - 渭}{蟽}\) For the lower score (a): \(z_a = \frac{300 - 500}{100} = -2\) For the upper score (b): \(z_b = \frac{550 - 500}{100} = 0.5\)
03

Find the area under the standard normal curve

Now, we need to find the area under the standard normal curve between z-scores -2 and 0.5. To do this, we will use a standard normal table (also known as z-table). First, find the area to the left of z_a=-2 in the z-table: P(Z< -2) 鈮� 0.0228 Next, find the area to the left of z_b=0.5 in the z-table: P(Z< 0.5) 鈮� 0.6915
04

Calculate the probability of scoring between 300 and 550

To find the probability of a randomly chosen test-taker scoring between 300 and 550, we need to find the area between these two z-scores. We can do this by subtracting the area to the left of z_a from the area to the left of z_b. Probability (P): P(a 鈮� X 鈮� b) = P(Z< 0.5) - P(Z< -2) P = 0.6915 - 0.0228 鈮� 0.6687 So, the probability that a randomly chosen test-taker will score between 300 and 550 is approximately 0.6687 or 66.87%.

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Key Concepts

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

z-score
The z-score is a statistical measure that helps indicate how many standard deviations a particular score is from the mean of a distribution. To compute the z-score, you use the formula: \[ z = \frac{x - \mu}{\sigma} \] where:
  • \(x\) is the value you are examining,
  • \(\mu\) is the mean of the distribution, and
  • \(\sigma\) is the standard deviation.

In our example of LSAT scores, to find the z-scores for scores of 300 and 550, we subtract the mean (500) from each score and divide by the standard deviation (100). The z-scores turn out to be -2 for 300 and 0.5 for 550. This tells us that 300 is two standard deviations below the mean, whereas 550 is half a standard deviation above the mean.
standard normal table
The standard normal table, often referred to as the z-table, is a mathematical table utilized to find the probability that a statistic is observed below, above, or between values in a standard normal distribution. This distribution is a special type of normal distribution with a mean of 0 and a standard deviation of 1. When working with the z-table:
  • Locate the z-score in the table to find the area under the curve to the left of that score.
  • For negative z-scores, the table gives the probability of falling below that score.
  • For positive z-scores, it does the same but starts from zero upwards.

In our scenario, we've found the probabilities associated with z-scores -2 and 0.5. The area to the left of -2 is approximately 0.0228, while for 0.5, it is around 0.6915, denoting the proportion of the population expected to have scores below these z-scores.
probability calculation
Probability calculation is essential in determining the likelihood of a given event occurring within a set distribution. Utilized frequently in statistics, it can help quantify uncertainties by calculating the "area" under the curve of a probability density function. For our LSAT score question, we need the probability that a student's score falls between two values (300 and 550). By:
  • Finding the z-scores for both values,
  • Looking up those z-scores in the standard normal table, and
  • Subtracting the area (probability) to the left of -2 from the area to the left of 0.5,
we find the probability. The subtraction gives us the probability between these scores: \[ P(Z< 0.5) - P(Z< -2) = 0.6915 - 0.0228 \approx 0.6687 \]This means there is approximately a 66.87% chance that a test-taker's score will fall between 300 and 550.
mean and standard deviation
Understanding the mean and standard deviation is crucial for interpreting any normal distribution effectively.
  • The **mean** (\(\mu\)) is the average value of a data set, representing its center. In our LSAT scores example, the mean is 500.
  • The **standard deviation** (\(\sigma\)) measures the amount of variation or dispersion from the mean. In this example, the standard deviation is 100, indicating the scores are spread out over a range of values.

These two parameters define the shape and spread of the normal distribution. The mean determines the location of the center, while the standard deviation specifies how much the data can deviate on average from the mean. When examining our problem, knowing that the scores are normally distributed with a mean of 500 and a standard deviation of 100 helps predict how scores deviate from the average, aiding us in calculating probabilities for scores within this distribution.

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Most popular questions from this chapter

Exercises \(45-50\) are based on the following information, gathered from student testing of a statistical software package called MODSTAT. \(^{56}\) Students were asked to complete certain tasks using the software, without any instructions. The results were as follows. (Assume that the time for each task is normally distributed. Find the probability that a student will take at least 10 minutes to complete Task 3 .

Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{20}\) $$ -100,400,-200,-500,200,-300,-200,900,-100,200 $$ Compute the mean and median of the given sample. Fill in the blank: There were as many days with a change in the Dow above \(\quad\) points as there were with changes below that.

Following is an excerpt from a full-page ad by MoveOn.org in the New York Times criticizing President G.W. Bush: \(^{33}\) On Tax Cuts: George Bush: "... Americans will keep, this year, an average of almost \(\$ 1,000\) more of their own money." The Truth: Nearly half of all taxpayers get less than \(\$ 100\). And \(31 \%\) of all taxpayers get nothing at all. The statements referred to as "The Truth" contradict the statement attributed to President Bush, right? Explain.

Exercises \(45-50\) are based on the following information, gathered from student testing of a statistical software package called MODSTAT. \(^{56}\) Students were asked to complete certain tasks using the software, without any instructions. The results were as follows. (Assume that the time for each task is normally distributed. It can be shown that if \(X\) and \(Y\) are independent normal random variables with means \(\mu_{X}\) and \(\mu_{Y}\) and standard deviations \(\sigma_{X}\) and \(\sigma_{Y}\) respectively, then their sum \(X+Y\) is also normally distributed and has mean \(\mu=\mu_{X}+\mu_{Y}\) and standard deviation \(\sigma=\sqrt{\sigma_{X}^{2}+\sigma_{Y}^{2}}\). Assuming that the time it takes a student to complete each task is independent of the others, find the probability that a student will take at least 20 minutes to complete both Tasks 1 and \(2 .\)

According to a study, \({ }^{31}\) the probability that a randomly selected teenager shopped at a mall at least once during a week was .63. How many teenagers in a randomly selected group of 40 would you expect to shop at a mall during the next week? HINT [See Example 5.]
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