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Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(0 \leq z \leq 1.62)$$

Short Answer

Expert verified
The probability \(P(0 \leq z \leq 1.62)\) is approximately 0.4474.

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution, often represented by the variable \(z\), is a normal distribution with a mean of 0 and a standard deviation of 1. This allows us to use standard normal distribution tables or calculators to find probabilities.
02

Identifying the Probability

We need to find the probability that \(z\) falls between 0 and 1.62. This is symbolized by \(P(0 \leq z \leq 1.62)\).
03

Using the Z-Table

To find \(P(0 \leq z \leq 1.62)\), we will look up the values in the Z-table. The Z-table gives the probability of a \(z\) value being less than a given number. First, find \(P(z \leq 1.62)\) using the Z-table.
04

Calculating P(z ≤ 1.62)

Using a Z-table, locate 1.6 in the leftmost column and 0.02 in the top row. The intersection provides the probability \(P(z \leq 1.62)\). According to most Z-tables, \(P(z \leq 1.62)\) is approximately 0.9474.
05

Calculating P(z ≤ 0)

By definition of the standard normal distribution, \(P(z \leq 0) = 0.5\), as the probability is evenly split at the mean (0).
06

Finding the Probability between 0 and 1.62

Finally, calculate \(P(0 \leq z \leq 1.62)\) by subtracting \(P(z \leq 0)\) from \(P(z \leq 1.62)\). This gives us \(0.9474 - 0.5 = 0.4474\).
07

Shading the Area

On a graph of the standard normal distribution, shade the area between \(z = 0\) and \(z = 1.62\). This area represents the probability \(P(0 \leq z \leq 1.62)\).

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Key Concepts

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

Z-Table
The Z-Table is an essential tool in statistics. It helps us find probabilities related to the standard normal distribution. The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. When we use a Z-table, we can determine how likely it is for a certain value of the random variable, often represented as \(z\), to occur.

The Z-Table works by providing the probability that the \(z\) score will be less than or equal to a particular value. Each entry in the Z-table corresponds to a specific \(z\) score and gives the cumulative probability up to that score. For example in our context, to find \(P(z \leq 1.62)\), we look for 1.6 in the left column and 0.02 at the top row of the Z-Table, and the intersecting value gives us the probability we need.
  • This table helps in problems like finding \(P(0 \leq z \leq 1.62)\), where multiple \(z\) scores are involved.
  • Cumulative means it adds up all probabilities up to that point.
Probability Calculation
The task of calculating probability in a standard normal distribution involves combining information from Z-tables. The main aim is to find the likelihood that a random variable \(z\) will land within a specified range, here between 0 and 1.62.

To find \(P(0 \leq z \leq 1.62)\):
  • First, use the Z-Table to determine \(P(z \leq 1.62)\). This gives the entire probability of \(z\) being less than or equal to 1.62.
  • Because the standard normal distribution is perfectly symmetrical about the mean of 0, the probability that \(z\) is less than or equal to 0 is always 0.5.
  • To get the probability between 0 and 1.62, subtract the left part, \(P(z \leq 0)\), from the total \(P(z \leq 1.62)\).
  • This gives us \(0.9474 - 0.5 = 0.4474\). Thus, \(P(0 \leq z \leq 1.62) = 0.4474\).
Shading Probability Areas
Visualizing problems with shading on the standard normal distribution curve helps in understanding the concepts better. When tasked with shading, it simply means marking the area under the curve that corresponds to the calculated probability.

For the probability \(P(0 \leq z \leq 1.62)\):
  • You would shade the region under the curve starting from \(z = 0\) and ending at \(z = 1.62\).
  • The shaded area visually represents the probability that \(z\) is greater than or equal to 0 and less than or equal to 1.62, which we've calculated as 0.4474.
  • This technique is useful because it provides a quick method to grasp how probabilities work in the standard normal distribution and what part of the data they represent.
Shading is a powerful visualization tool in statistics, aiding learners in connecting abstract numbers to real concepts.

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Most popular questions from this chapter

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \leq 0)$$

Airline Flights: No-Shows Based on long experience, an airline has found that about \(6 \%\) of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 267 ticket reservations for an airplane with only 255 seats. (a) What is the probability that a person holding a reservation will show up for the flight? (b) Let \(n=267\) represent the number of ticket reservations. Let \(r\) represent the number of people with reservations who show up for the flight. Which expression represents the probability that a seat will be available for everyone who shows up holding a reservation? $$P(255 \leq r) ; P(r \leq 255) ; P(r \leq 267) ; P(r=255)$$ (c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation?

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