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Chapter 7: Problem 40

Find the indicated probability of the standard normal random variable \(Z\). $$P(Z \geq-0.92)$$

Short Answer

Expert verified
0.8212

Step by step solution

01

- Understand the Problem

The problem requires finding the probability that the standard normal random variable (Z) is greater than or equal to -0.92, denoted as \(P(Z \geq -0.92)\). This involves using the standard normal distribution.
02

- Use the Z-table

Look up the value of -0.92 in the standard normal (Z) table. The Z-table typically gives the cumulative probability for a given Z-score from the left up to that Z-score.
03

- Find the Cumulative Probability for -0.92

Find the cumulative probability corresponding to \(Z = -0.92\) in the Z-table. This cumulative probability tells us the probability that Z is less than or equal to -0.92. From the Z-table, \(P(Z \leq -0.92) = 0.1788\).
04

- Use Complement to Find the Desired Probability

Since we need the probability that Z is greater than or equal to -0.92, we use the complement rule: \(P(Z \geq -0.92) = 1 - P(Z \leq -0.92)\).
05

- Calculate the Probability

Subtract the cumulative probability from 1: \(P(Z \geq -0.92) = 1 - 0.1788 = 0.8212\).

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Key Concepts

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

Z-score
The Z-score, also known as the standard score, helps in understanding where a particular value stands in relation to the mean of a distribution. It is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] Here, \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Z-scores allow us to compare different values from the same or different normal distributions by converting them to a common scale. A positive Z-score indicates a value above the mean, while a negative Z-score signifies a value below the mean. This standardization simplifies the calculation of probabilities and comparison of datasets.
Cumulative Probability
Cumulative probability is the probability that a random variable takes on a value less than or equal to a specific value. In the context of the standard normal distribution, the cumulative probability for a Z-score tells us the likelihood that a standard normal variable is less than or equal to that Z-score. When you look up a Z-score in the Z-table, the value you find is the cumulative probability. For instance, in our exercise, the cumulative probability for \( Z = -0.92 \) is 0.1788 which means there's a 17.88% chance that Z will take on a value less than or equal to -0.92.
Complement Rule
The complement rule is a fundamental concept in probability theory, stating that the probability of an event happening is equal to one minus the probability of it not happening. Mathematically, it is expressed as: \[ P(A) = 1 - P(A') \] where \(P(A)\) is the probability of event A, and \(P(A')\) is the probability of the complement event that A does not occur. In our exercise from the Z-table, we found that \( P(Z \leq -0.92) = 0.1788 \). Using the complement rule to find \( P(Z \geq -0.92) \), we get: \[ P(Z \geq -0.92) = 1 - P(Z \leq -0.92) = 1 - 0.1788 = 0.8212 \] Hence, there is an 82.12% chance that a standard normal variable is greater than or equal to -0.92.
Z-table
The Z-table is an essential tool used in statistics to find cumulative probabilities for standard normal distributions. It lists Z-scores and their corresponding cumulative probabilities. There are two common types of Z-tables:
  • One that provides the cumulative probability from the mean (0) to a positive Z-score.
  • Another that gives the cumulative probability from the left up to a given Z-score.
In our exercise, we used the second type of Z-table. Finding the Z-score of -0.92 in the table gives us a cumulative probability of 0.1788, denoting the area under the curve to the left of -0.92. This process simplifies many probability calculations by turning complex integrals into straightforward lookups.

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Most popular questions from this chapter

Steel rods are manufactured with a mean length of 25 centimeter \((\mathrm{cm}) .\) Because of variability in the maufacturing process, the lengths of the rods are approximateIy normally distributed with a standard deviation of \(0.07 \mathrm{cm} .\) (a) What proportion of rods has a length less than 24.9 \(\mathrm{cm} ?\) (b) Any rods that are shorter than \(24.85 \mathrm{cm}\) or longer than \(25.15 \mathrm{cm}\) are discarded. What proportion of rods will be discarded? (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between \(24.9 \mathrm{cm}\) and \(25.1 \mathrm{cm} ?\)

Assume the random variable \(X\) is normally distributed with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. $$P(56

In a 2003 study, the Accreditation Council for Graduate Medical Education found that medical residents' mean number of hours worked in a week is \(81.7 .\) Suppose the number of hours worked per week by medical residents is approximately normally distributed with a standard deviation of 6.9 hours. (Source: www.medrecinst.com) (a) Determine the 75 th percentile for the number of hours worked in a week by medical residents. (b) Determine the number of hours worked in a week that makes up the middle \(80 \%\) of medical residents.

Find the indicated \(Z\) -score. Be sure to draw a standard normal curve that depicts the solution. Find the \(Z\) -score such that the area under the standard normal curve to the left is 0.1.

Assume the random variable \(X\) is normally distributed with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. $$P(56 \leq X<66)$$
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