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The Medical College Admission Test. The Normal distribution with mean \(\mu=499.6\) and standard deviation \(\sigma=10.4\) is a good description of the total score on the Medical College Admission Test (MCAT). \({ }^{9}\) This is a continuous probability model for the score of a randomly chosen student. Call the score of a randomly chosen srudent \(X\) for short. (a) Write the event "the student chosen has a score of 510 or higher" in terms of \(\boldsymbol{K}\). (b) Find the probability of this event.

Short Answer

Expert verified
The probability a student scores 510 or higher is 0.1587.

Step by step solution

01

Define the Event in Terms of X

The problem asks us to express the event "the student chosen has a score of 510 or higher" using the random variable \( X \). Since \( X \) represents the score of a randomly chosen student, the event can be written as \( X \geq 510 \).
02

Standardize the Random Variable

To find the probability, we first convert the given value 510 to a standardized value using the formula for the z-score: \( z = \frac{X - \mu}{\sigma} \). Here, \( X = 510 \), \( \mu = 499.6 \), and \( \sigma = 10.4 \).\[ z = \frac{510 - 499.6}{10.4} \]Calculate this to find the z-score.
03

Calculate the Z-Score

Substitute the values into the z-score formula:\[ z = \frac{510 - 499.6}{10.4} = \frac{10.4}{10.4} = 1 \]The z-score corresponding to a score of 510 is 1.
04

Find the Probability Using Z-Table

Now, use a standard normal distribution table (z-table) to find the probability that \( Z \) is greater than or equal to 1. The z-table provides the area to the left of a given z-score. For \( z = 1 \), the table gives approximately 0.8413, which is the probability that \( Z \) is less than 1. To find \( P(Z \ge 1) \), subtract this value from 1:\[ P(Z \ge 1) = 1 - 0.8413 = 0.1587 \]Thus, \( P(X \ge 510) = 0.1587 \).

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Key Concepts

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

Medical College Admission Test
The Medical College Admission Test, or MCAT, is a standardized exam used in the United States and other countries to assess the readiness of students applying to medical school. The score on the MCAT is an important factor in the admission process. The MCAT is meant to test not only your knowledge of science but also your problem-solving and critical thinking skills. Because these skills are closely tied to success in medical school, having a good MCAT score can significantly boost your chances of being admitted. Understanding the distribution of MCAT scores can help identify how well a student performed relative to others. The scores typically follow a normal distribution, which means most examinees score near the average, with fewer people achieving low or high scores.
z-score
The z-score is a statistical measure that describes a value's position relative to the mean of a group of values. It indicates how many standard deviations an element is from the mean.To calculate a z-score, you use the formula: \[ z = \frac{X - \mu}{\sigma} \]- **X** is the value you're evaluating.- **\( \mu \)** is the mean of the group.- **\( \sigma \)** is the standard deviation of the group.In the context of our example with MCAT scores, we used a mean score of 499.6 and a standard deviation of 10.4 to determine the z-score for a student scoring 510. This allows us to assess how far above or below the average this score falls.
probability calculation
Probability calculation is a mathematical process used to determine the likelihood of a particular outcome. In the context of MCAT scores, once you have calculated a z-score, you can use probability tables to determine how likely it is for a student to score at or above this value.For a normal distribution, the z-table helps find the probability of a z-score falling to the left of the given value, which is useful for determining the likelihood of exceeding certain scores.With the example of an MCAT score of 510:- First, find the z-score.- Next, use the z-value to determine the probability.The calculation: \( P(Z \ge 1) = 1 - 0.8413 = 0.1587 \)This indicates an approximate 15.87% probability that a randomly chosen student scores 510 or higher.
standard deviation
Standard deviation is a crucial concept in statistics that measures the extent of variation or dispersion in a set of values. It is symbolized by \( \sigma \) when discussing a population.The larger the standard deviation, the more spread out the numbers are, compared to the mean.To calculate the standard deviation, you typically follow these steps:
  • Find the mean (average) of your data set.
  • Subtract the mean and square the result for each number.
  • Find the average of those squared differences.
  • Take the square root of that average to find the standard deviation.
In our MCAT example, the standard deviation of 10.4 implies that most students score within 10.4 points above or below the mean score of 499.6, providing a measure of variability in exam performance.

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Most popular questions from this chapter

Education among young adults. Choose a young adult (aged 25-29) at random. The probability is \(0.10\) that the person chosen did not complete high school, \(0.27\) that the person has a high school diploma but no further education, and \(0.34\) that the person has at least a bachelor's degree. (a) What must be the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor's degree? (b) What is the probability that a randomly chosen young adult has at least a high school education?

Land in Canada. Canada's national statistics agency, Statistics Canada, says that the land area of Canada is \(9,094,000\) square kilometers. Of this land, \(4,176,000\) square kilometers are forested. Choose a square kilometer of land in Canada at random. (a) What is the probability that the area you chose is forested? (b) What is the probability that it is not forested?

Running a Mile. A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean \(7.11\) minutes and standard deviation \(0.74\) minute. \({ }^{11}\) Choose a student at random from this group and call his time for the mile \(Y\). (a) Is \(Y\) a finite or continuous random variable? Explain your answer. (b) Say in words what the meaning of \(P(Y \geq 3.0)\) is. What is this probability? (c) Write the event "the student could run a mile in less than 6 minutes" in terms of values of the random variable \(Y\). What is the probability of this event?

Are They Disjoint? Which of the following pairs of events, \(A\) and \(B\), are disjoint? Explain your answers. (a) A person is selected at random. A is the event \({ }^{\alpha}\) sex of the person selected is male"; \(B\) is the event "sex of the person selected is female." (b) A person is selected at random. \(A\) is the event "the person selected earns more than \(\$ 100,000\) per year"; \(B\) is the event "the person selected earns more than \(\$ 250,000\) per year." (c) A pair of dice are tossed. \(A\) is the event "one of the dice is a \(3^{\prime \prime}, B\) is the event "the sum of the two dice is 3 ."

Survey accuracy. A sample survey contacted an SRS of 2854 registered voters shortly before the 2012 presidential election and asked respondents whom they planned to vote for. Election results show that \(51 \%\) of registered voters voted for Barack Obama. We will see later that in this situation the proportion of the sample who planned to vote for Barack Obama (call this proportion \(V\) ) has approximately the Normal distribution with mean \(\mu=0.51\) and standard deviation \(\sigma=0.009\). (a) If the respondents answer truthfully, what is \(P(0.49 \leq V \leq 0.53)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion \(0.51\) within plus or minus \(0.02\). (b) In fact, \(49 \%\) of the respondents said they planned to vote for Barack Obama \((V=0.49)\). If respondents answer truthfully, what is \(P(V \leq 0.49)\) ?

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