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X has a normal distribution with variance of \(25 .\) Given that the probability that \(x\) is less than 27.5 is \(0.312,\) find the mean \(\mu\) of this distribution.

Short Answer

Expert verified
The mean \(\mu\) is 29.95.

Step by step solution

01

Determine Standard Deviation

The variance of a normal distribution is given as \(25\). The standard deviation, \(\sigma\), is the square root of the variance. Therefore, \(\sigma = \sqrt{25} = 5\).
02

Use the Standard Normal Distribution

We use the z-score formula: \(z = \frac{X - \mu}{\sigma}\), where \(X = 27.5\), \(\sigma = 5\), and \(P(X < 27.5) = 0.312\). We need to find the corresponding z-value such that \(P(Z < z) = 0.312\).
03

Find the Z-value

Using a standard normal distribution table or calculator, find the z-value for a probability of \(0.312\). The closest value is approximately \(z = -0.49\).
04

Solve for the Mean \(\mu\)

Substitute the known values into the z-score formula: \(-0.49 = \frac{27.5 - \mu}{5}\). Solve for \(\mu\) by multiplying both sides by \(5\), which gives \(-2.45 = 27.5 - \mu\). Then, \(\mu = 27.5 + 2.45 = 29.95\).
05

Conclusion

The mean \(\mu\) of the normal distribution is \(29.95\).

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Key Concepts

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

Understanding Variance
Variance is a statistical measure that represents how much values in a dataset diverge from the mean. In a normal distribution, variance is denoted by \(\sigma^2\). It is essentially the average of the squared differences from the mean. A larger variance indicates that the data points are spread out over a wider range of values. However, finding variance alone might not give the best insight into the data's variability, which is why we often refer to standard deviation. For example, if a distribution has a variance of \(25\), it means on average, the squared deviations from the mean are \(25\). To further utilize this information, we take the square root of variance to find the standard deviation.
Unpacking Standard Deviation
Standard deviation, often denoted by \(\sigma\), is a key concept in statistics that describes the amount of variation or dispersion in a set of values. It is the square root of variance, \(\sqrt{\sigma^2}\), which makes it directly interpretable in the same units as the data.
  • It tells us how much the individual data points differ from the mean.
  • A small standard deviation indicates that data points tend to be close to the mean.
  • A large standard deviation suggests the opposite.
For instance, if the variance is \(25\), the standard deviation will be \(5\), indicating that the data points usually deviate by \(5\) units from the mean.
Explaining the Z-Score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations.
  • The z-score tells us how many standard deviations a data point is from the mean.
  • A positive z-score indicates the data point is above the mean, while a negative z-score indicates it's below the mean.
In calculations, the z-score is defined as \( z = \frac{X - \mu}{\sigma} \). In our example, if a data point \(X\) is \(27.5\), the z-score can be identified as \(-0.49\), which informs us that \(X\) is approximately half a standard deviation below the mean.
Mean Calculation Basics
Calculating the mean, especially within a normal distribution, is integral to determine central location. The mean, denoted by \(\mu\), summarizes the central tendency of a data set, essentially offering a point of balance.To isolate the mean, rearrange known equations, such as the z-score formula: \( z = \frac{X - \mu}{\sigma} \). Solving for \(\mu\) involves algebra to position \(\mu\) as the subject; for example, turning it into \( \mu = X - z \cdot \sigma \). In our scenario, the calculated mean \(\mu\) is \(29.95\), derived from setting \(X = 27.5\), \(\sigma = 5\), and \(z = -0.49\). This confirms that on average, data centers around \(29.95\) in this normal distribution.

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