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A variable is normally distributed with a mean of 120 and a standard deviation of \(5 .\) One score is randomly sampled. What is the probability it is above \(127 ?\)

Short Answer

Expert verified
The probability is approximately 8.08\%.

Step by step solution

01

Understand the Problem

We are given a normally distributed variable with mean \(\mu = 120\) and standard deviation \(\sigma = 5\). We need to find the probability that a randomly sampled score is greater than \(127\).
02

Calculate the Z-Score

The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For \(X = 127\), the Z-score is \( Z = \frac{127 - 120}{5} = 1.4 \).
03

Find the Probability Using the Z-Table

Using a standard normal distribution table (Z-table), we look up the probability corresponding to a Z-score of 1.4. The table provides the probability that a score is less than 1.4. This is approximately 0.9192.
04

Calculate the Probability Greater Than the Z-Score

Since we need the probability of the score being greater than \(127\), we calculate this as \(1 - P(Z < 1.4)\). Thus, \(1 - 0.9192 = 0.0808\).
05

Final Answer

The probability that a randomly sampled score from this distribution is greater than \(127\) is \(0.0808\), or approximately 8.08\%.

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Key Concepts

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

Z-score calculation
Calculating the Z-score is an essential skill when working with a normal distribution. A Z-score tells us how many standard deviations an individual data point is from the mean of the distribution. To calculate a Z-score, use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
In this formula, \(X\) is the individual data point, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. By standardizing the data point with this formula, you can compare it easily with other scores across different normal distributions.

For instance, consider if you want to determine how a score of 127 compares to a normal distribution with a mean (\(\mu)\) of 120 and a standard deviation (\(\sigma)\) of 5. The calculation goes as follows:
  • \( Z = \frac{127 - 120}{5} = 1.4 \)
This Z-score of 1.4 indicates that the score of 127 is 1.4 standard deviations above the mean.
probability calculation
Once you have calculated the Z-score, the next step is finding the probability associated with it. This process helps you understand the likelihood of a random sample score being above or below a certain value in a normal distribution.

For a Z-score of 1.4, you need to determine the probability that corresponds to this score in the context of the standard normal distribution.
  • First, look up the Z-score in a Z-table to find the probability \( P(Z < 1.4) \).
  • The table typically gives the cumulative probability of a score being less than your Z-score.
  • In our case, \( P(Z < 1.4) \) is approximately 0.9192, indicating a 91.92% chance that a score is less than 127.
However, the problem asks for the probability greater than this score. You calculate it by subtracting the cumulative probability from 1:
  • \( P(Z > 1.4) = 1 - P(Z < 1.4) = 0.0808 \)
Therefore, the probability of a randomly sampled score being greater than 127 is about 8.08%.
Z-table usage
A Z-table, also known as a standard normal distribution table, is a valuable tool for finding the probability associated with a specific Z-score. Here’s how you can effectively use it:
  • First, identify the Z-score you need. For example, a Z-score of 1.4.
  • Look for the value of 1.4 in the Z-table. The table usually displays cumulative probabilities from the left side of the distribution to your Z-score.
  • Find the corresponding probability that shows the likelihood of a score being below the Z-score. For a Z-score of 1.4, the cumulative probability is 0.9192.
When dealing with probabilities greater than the Z-score, subtract the table value from 1, as in this case, you want \( P(Z > 1.4) \), which is 0.0808.

Using the Z-table allows straightforward interpretation of normal distribution probabilities, making it a critical resource for anyone dealing with statistical data analysis.

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Most popular questions from this chapter

Suppose that combined verbal and math SAT scores follow a normal distribution with mean 896 and standard deviation \(174 .\) Suppose further that Peter finds out that he scored in the top \(3 \%\) of SAT scores. Determine how high Peter's score must have been.

A group of students at a school takes a history test. The distribution is normal with a mean of \(25,\) and a standard deviation of \(4 .\) (a) Everyone who scores in the top \(30 \%\) of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate? (b) The top \(5 \%\) of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?

A set of test scores are normally distributed. Their mean is 100 and standard deviation is \(20 .\) These scores are converted to standard normal z scores. What would be the mean and median of this distribution? a. 0 b. 1 c. 50 d. 100

True/false: In a normal distribution, \(11.5 \%\) of scores are greater than \(\mathrm{Z}=1.2 .\)

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of \(71 \mathrm{mph}\) and a standard deviation of \(8 \mathrm{mph}\). a. The current speed limit is \(65 \mathrm{mph}\). What is the proportion of vehicles less than or equal to the speed limit? b. What proportion of the vehicles would be going less than \(50 \mathrm{mph}\) ? c. A new speed limit will be initiated such that approximately \(10 \%\) of vehicles will be over the speed limit. What is the new speed limit based on this criterion? d. In what way do you think the actual distribution of speeds differs from a normal distribution?

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