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Find \(\operatorname{Pr}(-.47

Short Answer

Expert verified
The probability that a randomly selected value from the standard normal distribution lies between -0.47 and 94 is approximately \(0.6808\).

Step by step solution

01

Identify the given values

The given values are the lower limit \(-0.47\) and the upper limit \(94\).
02

Notation and Formula

We will use the notation \(\operatorname{Pr}(a<Z<b)\) for the probability that a standard normal value lies between a and b. To find this probability, we use the formula: \(\operatorname{Pr}(a<Z<b) = \operatorname{Pr}(Z < b) - \operatorname{Pr}(Z < a)\)
03

Finding probabilities from the Z-table

We will use the standard normal table (Z-table) to find the probabilities \(\operatorname{Pr}(Z < -0.47)\) and \(\operatorname{Pr}(Z < 94)\). From the Z-table, we find that: \(\operatorname{Pr}(Z < -0.47) = 0.3192\) Since \(94\) is very far from the mean (the mean is 0) and the distribution is continuous, we can assume that any value greater than 3 standard deviations is considered extremely unlikely. Hence we can safely say: \(\operatorname{Pr}(Z < 94) \approx 1\)
04

Calculate the probability in our interval

Plug the values into the formula: \(\operatorname{Pr}(-0.47 < Z < 94) = \operatorname{Pr}(Z < 94) - \operatorname{Pr}(Z < -0.47) = 1- 0.3192 = 0.6808\) Therefore, the probability that a randomly selected value from the standard normal distribution lies between -0.47 and 94 is approximately 0.6808.

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Key Concepts

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

Understanding the Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that serves as a foundational concept in statistics. It is defined by a bell-shaped curve that is symmetrical around the mean, which is zero. The only parameters that define it are its mean (\( \mu = 0 \)) and standard deviation (\( \sigma = 1 \)). This makes the standard normal distribution particularly useful for statisticians, as it allows them to standardize and compare different datasets easily.
The z-score is the number of standard deviations a data point is from the mean. With the standard normal distribution, these z-scores correspond to probabilities. This means we can easily calculate the probability of a value occurring within a certain range by using z-scores.
Because the standard normal distribution is a continuous distribution, any specific point on it has a probability of zero. Instead, we consider the probability of being less than or greater than a certain value, or within a certain range.
Using the Z-table
The Z-table is an essential tool for finding probabilities associated with the standard normal distribution. It contains cumulative probabilities for the standard normal distribution across a range of z-scores.
When you look up a value in the Z-table, you find the probability that a standard normal random variable is less than that particular z-score. This helps in determining the area under the curve to the left of that z-score.
For example, in our exercise, we looked up the values for \( Z = -0.47 \) and \( Z = 94 \).
  • \( \operatorname{Pr}(Z < -0.47) \) was approximately 0.3192, indicating that there's a 31.92% chance that a value falls below -0.47 in a standard normal distribution.
  • For a z-score as high as 94, which is unusually far to the right on the standard normal distribution, we assume \( \operatorname{Pr}(Z < 94) \) is nearly 1, representing almost all data under the curve.

The Z-table is incredibly handy when calculating probabilities, allowing us to deduce areas under the curve efficiently.
Calculating Normal Distribution Probability
Probability calculation in a normal distribution involves determining the likelihood of a random variable falling within a particular range. Utilizing the properties of the standard normal distribution simplifies this process.

To calculate the probability \( \operatorname{Pr}(a < Z < b) \), we follow this formula:\[ \operatorname{Pr}(a < Z < b) = \operatorname{Pr}(Z < b) - \operatorname{Pr}(Z < a) \]
Applying this formula requires values from the Z-table. In our example, we found the probability of \( Z < 94 \) and \( Z < -0.47 \), using those values to compute \( \operatorname{Pr}(-0.47 < Z < 94) \).
Subtracting the smaller cumulative probability from the larger one gives us the probability for the interval: \[ 1 - 0.3192 = 0.6808 \]
This resulting number, 0.6808, tells us there is a 68.08% chance that a randomly selected value from this distribution falls within \( -0.47 \) and \( 94 \).

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