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Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(P(|Z| \leq 2.50)\)

Short Answer

Expert verified
a) 0.4850, b) 0.3413, c) 0.4938, d) 0.9876, e) 0.9147, f) 0.9599, g) 0.9104, h) 0.0791, i) 0.0668, j) 0.9876.

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is denoted by the variable \(Z\). To solve problems involving \(Z\), we often use tables or software to find probabilities corresponding to \(Z\) values under the standard normal curve.
02

Visualizing the Problem

For each probability, sketch a graph of the standard normal distribution. Shade the area of interest under the curve to represent the probability we need to calculate. This helps in understanding which area under the curve we need to find.
03

Solving Problem a

To find \(P(0 \leq Z \leq 2.17)\), first recall that \(P(0 \leq Z \leq z) = P(Z \leq z) - P(Z \leq 0)\). From standard normal distribution tables, \(P(Z \leq 2.17) = 0.9850\) and \(P(Z \leq 0) = 0.5\). Thus, \(P(0 \leq Z \leq 2.17) = 0.9850 - 0.5 = 0.4850\).
04

Solving Problem b

For \(P(0 \leq Z \leq 1)\), we use a similar approach. \(P(Z \leq 1) = 0.8413\). Therefore, \(P(0 \leq Z \leq 1) = 0.8413 - 0.5 = 0.3413\).
05

Solving Problem c

To find \(P(-2.50 \leq Z \leq 0)\), we use the symmetry of the normal distribution: \(P(-z \leq Z \leq 0) = P(0 \leq Z \leq z)\). \(P(Z \leq -2.50) = 0.0062\), so \(P(-2.50 \leq Z \leq 0) = 0.5 - 0.0062 = 0.4938\).
06

Solving Problem d

The probability \(P(-2.50 \leq Z \leq 2.50)\) is found by \(P(Z \leq 2.50) - P(Z \leq -2.50)\). From tables, \(P(Z \leq 2.50) = 0.9938\) and \(P(Z \leq -2.50) = 0.0062\), so \(P(-2.50 \leq Z \leq 2.50) = 0.9938 - 0.0062 = 0.9876\).
07

Solving Problem e

To find \(P(Z \leq 1.37)\), use the table, which gives \(P(Z \leq 1.37) = 0.9147\).
08

Solving Problem f

For \(P(-1.75 \leq Z)\), find the complement: 1 minus the probability that \(Z\) is less than \(-1.75\). \(P(Z \leq -1.75) = 0.0401\). Thus, \(P(-1.75 \leq Z) = 1 - 0.0401 = 0.9599\).
09

Solving Problem g

To find \(P(-1.50 \leq Z \leq 2.00)\), calculate \(P(Z \leq 2.00)\) and \(P(Z \leq -1.50)\). \(P(Z \leq 2.00) = 0.9772\) and \(P(Z \leq -1.50) = 0.0668\); thus, \(P(-1.50 \leq Z \leq 2.00) = 0.9772 - 0.0668 = 0.9104\).
10

Solving Problem h

For \(P(1.37 \leq Z \leq 2.50)\), calculate \(P(Z \leq 2.50)\) and \(P(Z \leq 1.37)\). From the table, \(P(Z \leq 1.37) = 0.9147\) and \(P(Z \leq 2.50) = 0.9938\). Hence, \(P(1.37 \leq Z \leq 2.50) = 0.9938 - 0.9147 = 0.0791\).
11

Solving Problem i

To find \(P(Z \geq 1.50)\), calculate the complement: 1 minus the probability of \(Z\) being less than \(1.50\). \(P(Z \leq 1.50) = 0.9332\), so \(P(1.50 \leq Z) = 1 - 0.9332 = 0.0668\).
12

Solving Problem j

The probability \(P(|Z| \leq 2.50)\) refers to the probability of \(Z\) lying between \(-2.50\) and \(2.50\). We found earlier that \(P(-2.50 \leq Z \leq 2.50) = 0.9876\). Since this accounts for both directions, \(P(|Z| \leq 2.50) = 0.9876\).

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Key Concepts

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

Probability Calculation
Calculating probabilities with a standard normal distribution involves determining the area under the curve for certain values of the random variable. These calculations allow us to understand how likely it is for a random variable to fall within a specific range. Here's how it works:
  • First, identify the range of interest, such as from 0 to 2.17 or from -2.50 to 2.50.
  • Sketch a curve of the standard normal distribution and shade the area that corresponds to our range. This visualization helps you grasp which part of the curve represents the probability you need.
  • Use standard normal distribution tables or statistical software to find the probability values that correspond to the endpoint values of the range.
For example, to calculate the probability that a random variable falls between 0 and 2.17, we look up the probability that it is less than or equal to 2.17, then subtract the probability that it is less than or equal to 0. This gives us the probability for the range specifically.
Probability calculation often requires recognizing and utilizing the symmetry and properties of the normal distribution, such as the fact that the total area under the curve sums up to 1.
Normal Distribution Table
The normal distribution table, also known as the Z-table, is a helpful tool for finding the probabilities associated with a standard normal distribution. It typically provides the area under the curve to the left of a given Z-value.
  • The table comprises rows and columns representing the Z-values and the probabilities that a standard normal variable will be below that value.
  • By using this table, you can find out probabilities such as \( P(Z \leq 1.37) \) or \( P(Z \leq -2.50) \).
  • Remember that for positive Z-values, the table gives the probability \( P(Z \leq z) \), while for negative Z-values, you can use symmetry: \( P(Z \leq -z) = 1 - P(Z \geq z) \).
When you're asked to find probabilities for ranges, like \(-1.50 \leq Z \leq 2.00\), you calculate \( P(Z \leq 2.00) \) and subtract \( P(Z \leq -1.50) \). Using these tables becomes intuitive with practice, and they are crucial for many statistical analyses.
Random Variable Z
The random variable Z in the context of the standard normal distribution is essentially a variable that has been standardized. Here's what it means:
  • The mean of Z is 0, and the standard deviation is 1. This allows us to use the standard normal distribution table effectively.
  • Z-values are derived by taking the raw scores (X-values) of a variable and transforming them using the formula:
    \[ Z = \frac{X - \mu}{\sigma} \]
    where \( \mu \) is the mean and \( \sigma \) is the standard deviation.
  • These transformations allow us to compare scores from different normal distributions by converting them to a common scale.
Understanding Z as a standard normal random variable helps in calculating probabilities and making inferences about populations. Since any normal distribution can be transformed to a standard normal distribution, Z is a key component in statistical methods.

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