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Let \(z\) denote a variable that has a standard normal distribution. Determine each of the following probabilities: a. \(\quad P(z<2.36)\) b. \(\quad P(z \leq 2.36)\) c. \(\quad P(z<-1.23)\) d. \(\quad P(1.142)\) g. \(\quad P(z \geq-3.38)\) h. \(P(z<4.98)\)

Short Answer

Expert verified
a. 0.9906, b. 0.9906, c. 0.1093, d. 0.1226, e. 0.0706, f. 0.0228, g. 0.9996, h. 1.00

Step by step solution

01

- Understand the problem

We have a variable \(z\) with standard normal distribution. For each part a, b, ..., h, we have to find the probable values of \(z\) under specific conditions. We'll use the standard normal distribution table (Z-table) for this purpose. The total area under the curve of a standard normal distribution is 1, often divided by two sides: 0.5 to the left of 0 (mean) and 0.5 to the right.
02

- Solve Part a (\(P(z

This represents the probability that \(z\) is less than 2.36. Since we have \(z < 2.36\), we look directly at the Z-table. The value corresponding to 2.36 on the Z-table approximately equal to 0.9906.
03

- Solve Part b (\(P(z \leq 2.36)\))

The interpretation of the expression \(P(z \leq 2.36)\) is exactly the same as in step 2. The value from Z-table still applies, so the probability here is also approximately equal to 0.9906.
04

- Solve Part c (\(P(z

This represents the probability that \(z < -1.23\), that is \(z\) is less than -1.23. Probability for negative Z-scores is obtained by using symmetry property of standard normal distribution. Value corresponding to the Z-score of 1.23 from the Z-table is approximately 0.8907. Hence, the probability \(P(z < -1.23) = 0.5 - 0.8907 = 0.1093.\)
05

- Solve Part d (\(P(1.14

The probability for an interval can be determined by subtracting the probability of the lower Z-score from the probability of the upper Z-score. Hence, \(P(1.14 < z < 3.35) = P(z<3.35) - P(z<1.14)\). The probability corresponding to 3.35 and 1.14 from the Z-table is approximately 0.9996 and 0.8770 respectively. Hence, \(P(1.14 < z < 3.35) = 0.9996 - 0.8770 = 0.1226.\)
06

- Solve Part e (\(-0.77 \leq z \leq-0.55)\))

This represents the probability of \(z\) being in the given interval. Like the previous step, it's obtained by subtraction. \(P(-0.77 \leq z \leq-0.55) = P(z \leq -0.55) - P(z \leq -0.77)\). Using the Z-table, the probability corresponding to 0.55 and 0.77 is approximately 0.7088 and 0.7794. Since these are negative Z-scores, \(P(z \leq -0.55) = 0.5 - 0.7088 = 0.2912\) and \(P(z \leq -0.77) = 0.5 - 0.7794 = 0.2206\). So, \(P(-0.77 \leq z \leq-0.55) = 0.2912 - 0.2206 = 0.0706.\)
07

- Solve Part f (\(P(z>2)\))

The probability that \(z > 2\) is obtained by subtracting the value for a Z-Score of 2 from 1. The probability corresponding to 2 from the Z-table is approx. 0.9772 so, \(P(z > 2) = 1 - 0.9772 = 0.0228.\)
08

- Solve Part g (\(P(z \geq-3.38)\))

This means the probability that \(z\) lies from -3.38 to positive infinity. To find \(P(z \geq -3.38)\), look up 3.38 from the Z-table, which is approx. 0.9996. Using symmetry property, \(P(z \geq -3.38) = 1 - (0.5 - 0.9996) = 1 - 0.0004 = 0.9996.\)
09

- Solve Part h (\(P(z

This represents the probability that \(z < 4.98\). Whenever a z-score is greater than 3 or less than -3, it can be safely assumed that all areas under the curve will be accounted for. Thus, \(P(z<4.98)\) is approx. 1.00.

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Key Concepts

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

Understanding Probability in the Context of Standard Normal Distribution
In the realm of statistics, probability plays a crucial role. It's about understanding the likelihood that a particular event will occur. When we talk about probability in the context of a standard normal distribution, we're dealing with a bell-shaped curve that's symmetric around the mean of zero. The total area under the curve is 1, which represents 100% probability. This means any area under the curve corresponds to a probability less than or equal to 1. For instance, when we compute the probability that a variable is less than a particular value (like in part a, where we determine $P(z<2.36)$), we look at the cumulative area under the curve to the left of that point. This helps us understand the probability of that event happening. In simple terms, probability is all about how likely it is to find a value less than or equal to a specific number on the standard normal distribution.
The Role of Z-Score in Standard Normal Distribution
The z-score is a measure that describes a value's position relative to the mean of a group of values, measured in terms of standard deviations from the mean. In a standard normal distribution, we use z-scores to determine how far away a particular value is from the mean, which is zero. For example, if you have a z-score of 2, it means that the value is two standard deviations away from the mean. The z-score allows us to calculate the probability of a value occurring within a standard normal distribution by converting continuous data into a standardized form that can be easily interpreted. It simplifies the analysis of distributions, providing a clear way to compute probabilities and compare different datasets. In part f, you find the probability of a z-score being greater than 2, which involves understanding the position of this score on the curve.
How to Use the Z-Table
The Z-table, or the standard normal distribution table, is a mathematical tool that helps transform z-scores into probabilities. This table provides the probabilities associated with each z-score, specifying the area under the curve to the left of a given z-score. To find a given z-score on the Z-table, follow these steps:
  • First, determine the z-score you need to find from your problem. For example, for $P(z < 2.36)$, the z-score is 2.36.
  • Next, locate the whole number and the first decimal place of the z-score along the left-side column (e.g., 2.3 for a z-score of 2.36).
  • Then find the second decimal place on the top row (e.g., 0.06 for a z-score of 2.36).
  • Find the value at the intersection of the row and column you've identified. This value represents the probability you're interested in.
The Z-table helps students quickly and efficiently find probabilities, as seen in calculating $P(z < 2.36)$.
Understanding the Symmetry Property of Standard Normal Distribution
The symmetry property is a key feature of the standard normal distribution. This distribution is symmetrical around its mean of zero. It means that the probability of a z-score falling below a negative value is the same as it falling above the corresponding positive value. For example, if you know the probability of $P(z<1.23)$, you can easily find $P(z>-1.23)$ by using symmetry. This property implies that half of the data falls below the mean and half above, making it easy to calculate probabilities for negative z-scores. In part c of the solution, this concept is used to determine the probability for a z-score of -1.23 by first finding the probability for its positive counterpart. Understanding symmetry simplifies calculations and provides insights into the nature of the distribution, helping students solve similar standard normal distribution problems with greater ease.

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