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Suppose that IQs of East State University's students can be described by a Normal model with mean 130 and standard deviation 8 points. Also suppose that IQs of students from West State University can be described by a Normal model with mean 120 and standard deviation \(10 .\) a) We select a student at random from East State. Find the probability that this student's IQ is at least 125 points. b) We select a student at random from each school. Find the probability that the East State student's IQ is at least 5 points higher than the West State student's IQ. c) We select 3 West State students at random. Find the probability that this group's average IQ is at least 125 points. d) We also select 3 East State students at random. What's the probability that their average IQ is at least 5 points higher than the average for the 3 West Staters?

Short Answer

Expert verified
a) 73.6% b) 65.1% c) 0.47% d) 79.9%

Step by step solution

01

Analyze the Probability for East State Student's IQ

To find the probability that an East State student's IQ is at least 125, we'll use the Normal Distribution model. We first calculate the Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X = 125 \), \( \mu = 130 \), and \( \sigma = 8 \). Thus, \( Z = \frac{125 - 130}{8} = -0.625 \). Next, we look up the Z-score in the standard normal distribution table, which gives us the probability of \( P(Z < -0.625) \). Since we need \( P(X \geq 125) \), we calculate \( 1 - P(Z < -0.625) \).
02

Probability of Higher IQ from East State Student

For the probability that an East State student has an IQ at least 5 points higher than a West State student, we define the difference of the two IQs as a new variable \( D = X_E - X_W \). Since both are normally distributed, \( D \) is also normally distributed with \( \mu_D = 130 - 120 = 10 \) and \( \sigma_D = \sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.81 \). We find the probability that \( D \geq 5 \) by calculating the Z-score \( Z = \frac{5 - 10}{12.81} \approx -0.39 \) and then looking up this Z-score to find the corresponding probability.
03

Average IQ of Three West State Students

When selecting 3 West State students, the average IQ follows a normal distribution with \( \mu = 120 \) and \( \sigma = \frac{10}{\sqrt{3}} \). We want to find \( P(\bar{X} \geq 125) \). Using the Z-score formula for the sample mean, \( Z = \frac{125 - 120}{\frac{10}{\sqrt{3}}} \), we find \( Z \approx 2.60 \). We then find the probability from the Z-table.
04

Probability of Higher Average for East State Group

For finding the probability that the average IQ of 3 East State students is at least 5 points higher than that of 3 West State students, we consider \( D_3 = \bar{X}_E - \bar{X}_W \). The mean of \( D_3 \) is \( \mu_{D_3} = 130 - 120 = 10 \) and its standard deviation is \( \sigma_{D_3} = \sqrt{\frac{8^2}{3} + \frac{10^2}{3}} \approx 5.98 \). We find \( P(D_3 \geq 5) \) using \( Z = \frac{5 - 10}{5.98} \approx -0.84 \). We find the probability from the Z-table.

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Key Concepts

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

Z-Score
A z-score is a way to describe how far away a particular measurement is from the mean in terms of standard deviations. It is a crucial concept when dealing with normal distribution, as it helps in understanding where a value stands in a distribution. The formula for calculating a z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \( X \) is the value for which you are calculating the z-score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In our exercise, when asked to find the probability of an East State student's IQ being at least 125, we calculated the z-score to determine how 125 compares to the average IQ of that university. By finding the z-score, we identified how unusual a score of 125 is compared to the average of 130. The z-score calculation was \( -0.625 \), which showed that 125 is slightly below the mean. Understanding this, we could then use statistical tables to determine the likelihood of an IQ being at or above this score.
Mean and Standard Deviation
The mean and standard deviation are two statistical metrics used to summarize and understand data. The mean, often referred to as the average, indicates the central value in a dataset, while the standard deviation measures the amount of variation or dispersion from the mean.

  • The mean \( (\mu) \) provides insight into the central tendency of a distribution. In our example, East State has a mean IQ of 130, while West State's is 120.
  • The standard deviation \( (\sigma) \) gives an idea about the spread of the data - how much individual data points differ from the mean. East State has a standard deviation of 8, indicating that most student IQs are relatively close to the mean. On the other hand, West State's larger standard deviation of 10 suggests more variability in student IQs.
These metrics are pivotal when we calculate probabilities like those in our exercise, as they define the specific normal distribution model to which the data conforms.
Probability Calculation
Probability calculation involves determining the likelihood of a particular event occurring within a specific model. When dealing with normal distributions, probability calculations often use the z-score method to leverage standardized statistical tables.

In our exercise, we saw various scenarios where probabilities were calculated. For instance, to find the chance of a West State student's average IQ being at least 125, we calculated a z-score for the average, then used this score to find a probability from a normal distribution table. This method bridges our initial real-world problem with statistical theory by standardizing data through z-scores, allowing us to find probabilities efficiently.

Remember, calculating these probabilities allows for predicting outcomes based on an assumed distribution, measuring how likely it is that a random variable will fall within a particular range.
Random Sampling
Random sampling is a technique where each member of a population has an equal chance of being chosen to be part of the sample. It's essential for ensuring that the sample represents the population well, which is crucial for creating valid and reliable statistical inferences.

In our scenarios, students were chosen at random to calculate the probability of various IQ-related outcomes. Random sampling ensures that results aren't biased or skewed by non-representative data.
  • When selecting students at random, each student within either East or West State has the same likelihood of being chosen.
  • It’s fundamentally important for obtaining groups that reflect the true demographics and characteristics of the entire student body.
This method helps in applying probability models accurately, as it is assumed that the samples obtained reflect the actual distributions of the entire populations in the respective universities.

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Most popular questions from this chapter

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