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Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities.$$P(x \geq 120) ; \mu=100 ; \sigma=15$$

Short Answer

Expert verified
The probability is approximately 0.0918.

Step by step solution

01

Identify the Given Parameters

The problem provides us with the mean, \(\mu = 100\), and the standard deviation, \(\sigma = 15\). We are asked to find the probability that \(x\) is greater than or equal to 120, \(P(x \geq 120)\).
02

Convert to Z-Score

We need to convert the given \(x\) value to a standard normal variable \(z\). The formula to do this is \(z = \frac{x - \mu}{\sigma}\).Substituting the given values, we get \(z = \frac{120 - 100}{15} = \frac{20}{15} \approx 1.33\).
03

Use the Z-Table or Z-Calculator

We need to find the probability that \(z\) is greater than or equal to 1.33, denoted as \(P(z \geq 1.33)\). Look up this value in the Z-table or use a calculator that provides the cumulative probability for a standard normal distribution.
04

Calculate the Complementary Probability

The Z-table gives us the probability \(P(z \leq 1.33)\). Typically, the Z-table provides the area to the left of the z-score.The probability we need is \(P(z \geq 1.33)\). Calculate this by finding the complement: \(P(z \geq 1.33) = 1 - P(z \leq 1.33)\).From the Z-table, \(P(z \leq 1.33) \approx 0.9082\), thus \(P(z \geq 1.33) = 1 - 0.9082 \approx 0.0918\).
05

State the Conclusion

The probability that \(x\) is greater than or equal to 120 is approximately \(0.0918\).

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Key Concepts

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

Understanding the Z-Score
The Z-Score is a measure that describes how many standard deviations an element is from the mean of a distribution. It's a way of standardizing any normal distribution to compare it against the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

To calculate a Z-Score, use the formula:
  • \[ z = \frac{x - \mu}{\sigma} \]
Here, \( x \) is the value we're examining, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

When you compute the Z-Score for a value, you're naturally transforming it into a universal language. It allows you to determine the probability of scores occurring within a standard normal distribution. In essence, through the Z-Score, different data sets can be compared on the same scale.
Cumulative Probability Explained
Cumulative probability refers to the probability that a random variable, such as a test score or measurement, is less than or equal to a given value. It is a running total of probabilities.

In a graph of a normal distribution, the cumulative probability indicates the probability "up to" a certain point. It represents the area under the curve of the distribution to the left of or equal to a specific value.

For example, when we say \( P(z \leq 1.33) \), it signifies that we're calculating the total probability from the far left up to the point where \( z \) equals 1.33 on the z-axis. This helps in pinpointing how likely it is to observe a value within that particular range.
The Standard Normal Distribution
The Standard Normal Distribution is a special case of a normal distribution, defined with a mean of 0 and a standard deviation of 1. Represented by the variable \( z \), this distribution is used as a reference to calculate probabilities for any normal distribution.

One key feature is its bell shape and its symmetry around the mean, which facilitates modeling many naturally occurring datasets.
  • The total area under the curve is 1, which means it represents a total probability of 100%.
  • Approximately 68% of the data lies within one standard deviation of the mean.
  • About 95% falls within two standard deviations.
Because of its properties, the standard normal distribution is utilized to look up probabilities in a Z-table, which relates individual Z-Scores to their respective cumulative probabilities.
How Complementary Probability Works
Complementary probability deals with scenarios where you need to find the probability of the opposite event happening. When an event's probability is known, its complement is calculated by subtracting that probability from 1. This is because the total probability of all possible outcomes of any event is always 1.

So if you have a cumulative probability, say \( P(z \leq 1.33) = 0.9082 \), and you wish to find the probability of the opposite event (\( P(z \geq 1.33) \)), you employ the complementary probability method:
  • \[ P(z \geq 1.33) = 1 - P(z \leq 1.33) \]
  • This equates to \( P(z \geq 1.33) = 1 - 0.9082 = 0.0918 \) in our case.
This simple yet powerful tool helps deal with problems where direct probability measures aren't immediately available.

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Most popular questions from this chapter

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the right of \(z=1.52\)

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(-0.45 \leq z \leq 2.73)$$

Measurement errors from instruments are often modeled using the uniform distribution (see Problem 16 ). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds. (Reference: J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors," Journal of the Acoustical Society of America, Vol. 75, No. 1, pp. 197-201.) What is the probability that such measurements will be in error by (a) less than \(+0.03 \text { microsecond (i.e., }-0.05 \leq x<0.03) ?\) (b) more than -0.02 microsecond? (c) between -0.04 and +0.01 microsecond? (d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the right of \(z=-1.22\)

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \leq 0)$$

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