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Let \(Z\) be a standard normal random variable. Calculate (a) \(\operatorname{Pr}(-1.3 \leq Z \leq 0)\) (b) \(\operatorname{Pr}(.25 \leq Z)\) (c) \(\operatorname{Pr}(-1 \leq Z \leq 2.5)\) (d) \(\operatorname{Pr}(Z \leq 2)\)

Short Answer

Expert verified
a) 0.4032, b) 0.4013, c) 0.8351, d) 0.9772

Step by step solution

01

Understanding the Standard Normal Distribution Table

Values of the standard normal random variable, denoted by Z, correspond to the area under the curve of the standard normal distribution. The table provides the cumulative probability from the far left up to the Z value. We will use this to find the probabilities for the given ranges.
02

Calculate \(\text{Pr}(-1.3 \leq Z \leq 0)\)

Find the cumulative probability for Z = 0, which is given directly in the table as 0.5. Next, find the cumulative probability for Z = -1.3. According to the table, \(\text{Pr}(Z \leq -1.3) = 0.0968\). Subtract the cumulative probability for Z = -1.3 from that for Z = 0: \[\text{Pr}(-1.3 \leq Z \leq 0) = 0.5 - 0.0968 = 0.4032\].
03

Calculate \(\text{Pr}(0.25 \leq Z)\)

Find the cumulative probability for Z = 0.25 from the table, which is \(\text{Pr}(Z \leq 0.25) = 0.5987\). To find \(\text{Pr}(0.25 \leq Z)\), compute \[1 - \text{Pr}(Z \leq 0.25) = 1 - 0.5987 = 0.4013\].
04

Calculate \(\text{Pr}(-1 \leq Z \leq 2.5)\)

First, find the cumulative probability for Z = 2.5, which is \(\text{Pr}(Z \leq 2.5) = 0.9938\). Next, find the cumulative probability for Z = -1, which is \(\text{Pr}(Z \leq -1) = 0.1587\). Subtract the cumulative probability for Z = -1 from that for Z = 2.5: \[\text{Pr}(-1 \leq Z \leq 2.5) = 0.9938 - 0.1587 = 0.8351\].
05

Calculate \(\text{Pr}(Z \leq 2)\)

Find the cumulative probability for Z = 2 directly from the table, which is \(\text{Pr}(Z \leq 2) = 0.9772\). No further calculation is necessary.

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Key Concepts

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

Cumulative Probability
Cumulative probability represents the total probability that a random variable is less than or equal to a certain value. When dealing with a normal distribution, we often refer to the cumulative distribution function (CDF). The CDF gives us the area under the curve of the distribution to the left of a particular value.
For example, if you want to know the probability that a standard normal random variable (Z) is less than or equal to 1.5, you would use the Z table to find the cumulative probability up to 1.5. This value tells you the likelihood that your variable falls within that range.
Understanding cumulative probabilities is essential when performing probability calculations, as it allows you to determine the portion of data you are interested in.
Z-Score
The Z-score represents the number of standard deviations a given data point is from the mean of the distribution. For the standard normal distribution, the mean is 0 and the standard deviation is 1.
The Z-score helps convert any normal distribution to the standard normal distribution, making probability calculations and comparisons possible.
To calculate a Z-score, you use the formula: \(Z = \frac{(X - μ)}{σ}\), where X is the value, μ is the mean, and σ is the standard deviation.
For example, if you are given a value of 20, a mean of 10, and a standard deviation of 5, the Z-score is \(Z = \frac{(20 - 10)}{5} = 2\).
This means the value of 20 is 2 standard deviations above the mean.
Normal Distribution Table
The standard normal distribution table, or Z-table, provides cumulative probabilities for Z-scores in a standard normal distribution. It typically shows the probability that a standard normal variable is less than or equal to a given Z-score.
When using the Z-table:
  • Find the Z-score row corresponding to the integer part and the first decimal place of your Z-score.
  • Find the column that matches the second decimal place of your Z-score.
  • Locate the value at the intersection of your selected row and column.

For example, to find \(P(Z \leq 1.34)\), locate the row for 1.3 and the column for 0.04. The intersection gives you the cumulative probability.
The Z-table is a fundamental tool in statistics for finding cumulative probabilities quickly and accurately.
Probability Calculations
Probability calculations with the standard normal distribution involve using the Z-table to find cumulative probabilities and then performing basic arithmetic operations.
  • To find \(\text{Pr}(-1.3 \leq Z \leq 0)\), you first find \(\text{Pr}(Z \leq 0) = 0.5\) and \(\text{Pr}(Z \leq -1.3) = 0.0968\). Then, subtract to get the desired probability: \(0.5 - 0.0968 = 0.4032\).
  • To find \(\text{Pr}(0.25 \leq Z)\), find \(\text{Pr}(Z \leq 0.25) = 0.5987\), then subtract it from 1: \(1 - 0.5987 = 0.4013\).
  • To find \(\text{Pr}(-1 \leq Z \leq 2.5)\), find \(\text{Pr}(Z \leq 2.5) = 0.9938\) and \(\text{Pr}(Z \leq -1) = 0.1587\), and then subtract: \(0.9938 - 0.1587 = 0.8351\).

These calculations help understand the likelihood of a random variable falling within specified ranges.

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Most popular questions from this chapter

During a certain part of the day, the time between arrivals of automobiles at the tollgate on a turnpike is an exponential random variable with an expected value of 20 seconds. Find the probability that the time between successive arrivals is more than 60 seconds.

The quality-control department at a sewing machine factory has determined that 1 out of 40 machines does not pass inspection. Let \(X\) be the number of machines on an assembly line that pass inspection before a machine is found that fails inspection. (a) Write the formula for \(\operatorname{Pr}(X=n).\) (b) What is the probability that, of five machines coming oft the assembly line, the first four pass inspection and the fifth does not?

Suppose that the lifetime \(X\) (in hours) of a certain type of flashlight battery is a random variable on the interval \(30 \leq x \leq 50\) with density function \(f(x)=\frac{1}{20}\) \(30 \leq x \leq 50 .\) Find the probability that a battery selected at random will last at least 35 hours.

Normal Distribution and Life of a Tire Suppose that the life span of a certain automobile tire is normally distributed, with \(\mu=25,000\) miles and \(\sigma=2000\) miles. (a) Find the probability that a tire will last between 28,000 and 30,000 miles. (b) Find the probability that a tire will last more than 29,000 miles.

Let \(X\) be a continuous random variable with density function \(f(x)=3 x^{-4}, x \geq 1 .\) Compute \(\mathrm{E}(X)\) and \(\operatorname{Var}(X).\) If \(X\) is a random variable with density function \(f(x)\) on \(A \leq x \leq B\) the median of \(X\) is that number \(M\) such that $$\int_{A}^{M} f(x) d x=\frac{1}{2}.$$ In other words, \(\operatorname{Pr}(X \leq M)=\frac{1}{2}.\)

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