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Chapter 7: Problem 36

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

Short Answer

Expert verified
0.1788

Step by step solution

01

Understand the Problem

The problem requires finding the probability that the standard normal random variable is greater than 0.92. This is denoted as \(P(Z > 0.92)\).
02

Refer to Z-Table

A standard normal distribution table, or Z-table, shows the probability that a standard normal random variable \(Z\) is less than or equal to a given value \(z\) (\(P(Z \leq z)\)).
03

Find \(P(Z \leq 0.92)\)

Locate the value 0.92 on the Z-table. The value of \(P(Z \leq 0.92)\) from the Z-table is approximately 0.8212.
04

Calculate \(P(Z > 0.92)\)

To find \(P(Z > 0.92)\), subtract the value obtained from the Z-table from 1: \[ P(Z > 0.92) = 1 - P(Z \leq 0.92) \] Substitute the known value: \[ P(Z > 0.92) = 1 - 0.8212 \] Perform the subtraction: \[ P(Z > 0.92) = 0.1788 \]

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Key Concepts

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

Z-table
A Z-table is a crucial tool in statistics. It lets you find the probability that a standard normal random variable is less than or equal to a given value. This is written as (\(P(Z \leq z)\)).
Z-tables are organized by rows and columns to make looking up probabilities quick. The leftmost column and top row specify the value of Z to two decimal places. The numbers inside the table show the probability for those Z values.
For example, if we look up 0.92, we find that the probability (\(P(Z \leq 0.92)\)) is 0.8212. This means there's an 82.12% chance a standard normal random variable will be less than or equal to 0.92.
Standard normal random variable
In statistics, a standard normal random variable is a special type of random variable. It's denoted by the letter Z.
It follows a standard normal distribution. This means it has a mean (\(\mu)\)) of 0 and a standard deviation (\(\sigma)\)) of 1.
The distribution is symmetric around the mean. This makes it very useful for calculating probabilities.
The Z value helps to understand how far a point is from the mean in standard deviation units.
Probability calculation
Calculating probability with a Z-table involves several steps. First, identify the Z value you're interested in.
Then, find this value in the Z-table to get the cumulative probability (\(P(Z \leq z)\)).
If you need the probability of being greater than a Z value, you subtract the Z-table result from 1.
For example, to find (\(P(Z>0.92)\)):
  • Look up (\(P(Z \leq 0.92)\)) in the Z-table, which is 0.8212.
  • Subtract this from 1: (\(1 - 0.8212 = 0.1788)\)).
The result, 0.1788, is the probability that Z is greater than 0.92.
Z value
A Z value is a point on the standard normal distribution. It measures how many standard deviations a point is from the mean (0).
Positive Z values are above the mean, while negative Z values are below it.
For instance, a Z value of 0.92 indicates a point 0.92 standard deviations above the mean.
By looking up this Z value in a Z-table, we can find out the probability that a standard normal random variable is less than or equal to this value.
This is vital for various statistical calculations, including hypothesis testing and confidence intervals.

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Most popular questions from this chapter

Compute \(P(x)\) using the binomial probability formula. Then determine whether the normal distribution can be used as an approximation for the binomial distribution. If so, approximate \(P(x)\) and compare the result to the exact probability. $$n=60, p=0.4, X=20$$

A random variable \(X\) is normally distributed with \(\mu=25\) and \(\sigma=6\) (a) Compute \(Z_{1}=\frac{X_{1}-\mu}{\sigma}\) for \(X_{1}=18\) (b) Compute \(Z_{2}=\frac{X_{2}-\mu}{\sigma}\) for \(X_{2}=30\) (c) The area under the normal curve between \(X_{1}=18\) and \(X_{2}=30\) is \(0.6760 .\) What is the area between\(Z_{1}\)and \(Z_{2} ?\)

Find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the left of (a) \(Z=-3.49\) (b) \(Z=-1.99\) (c) \(Z=0.92\) (d) \(Z=2.90\)

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. A random sample of weekly work logs at an automobile repair station was obtained and the average number of customers per day was recorded. $$\begin{array}{lllll}26 & 24 & 22 & 25 & 23 \\\\\hline 24 & 25 & 23 & 25 & 22 \\\\\hline 21 & 26 & 24 & 23 & 24 \\ \hline 25 & 24 & 25 & 24 & 25 \\\\\hline 26 & 21 & 22 & 24 & 24\end{array}$$

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 0.2
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