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Chapter 6: Problem 12

Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(0 \leq z \leq .83)\) b. \(\quad P(-1.57 \leq z \leq 0)\) c. \(\quad P(z>.44)\) d. \(P(z \geq-.23)\) e. \(\quad P(z<1.20)\) f. \(\quad P(z \leq-.71)\)

Short Answer

Expert verified
a) 0.2967, b) 0.4418, c) 0.3300, d) 0.5910, e) 0.8849, f) 0.2389.

Step by step solution

01

Understanding Standard Normal Distribution

The standard normal distribution, or Z-distribution, is a normal distribution with a mean of 0 and a standard deviation of 1. Probabilities associated with specific Z-scores can be found using a Z-table or a standard normal distribution calculator.
02

Solving Part (a): Compute \(P(0 \leq z \leq 0.83)\)

We look up the Z-table or use a calculator to find \(P(z \leq 0.83)\). This value is approximately 0.7967. Since \(P(z \leq 0) = 0.5\), we find \(P(0 \leq z \leq 0.83) = P(z \leq 0.83) - P(z \leq 0) = 0.7967 - 0.5 = 0.2967\).
03

Solving Part (b): Compute \(P(-1.57 \leq z \leq 0)\)

Find \(P(z \leq 0) = 0.5\) and \(P(z \leq -1.57)\). Look up \(z = -1.57\) in the Z-table, which gives approximately 0.0582. Thus, \(P(-1.57 \leq z \leq 0) = P(z \leq 0) - P(z \leq -1.57) = 0.5 - 0.0582 = 0.4418\).
04

Solving Part (c): Compute \(P(z > 0.44)\)

Find \(P(z \leq 0.44)\) using the Z-table, which is approximately 0.6700. Therefore, \(P(z > 0.44) = 1 - P(z \leq 0.44) = 1 - 0.6700 = 0.3300\).
05

Solving Part (d): Compute \(P(z \geq -0.23)\)

First, calculate \(P(z < -0.23)\) using the Z-table, which is approximately 0.4090. Since \(P(z \geq -0.23) = 1 - P(z < -0.23) = 1 - 0.4090 = 0.5910\).
06

Solving Part (e): Compute \(P(z < 1.20)\)

Use the Z-table to find \(P(z \leq 1.20)\), which is approximately 0.8849. Because the inequality uses '<', it is equivalent to \(P(z \leq 1.20) = 0.8849\).
07

Solving Part (f): Compute \(P(z \leq -0.71)\)

Check the Z-table for \(z = -0.71\), yielding approximately 0.2389. Thus, \(P(z \leq -0.71) = 0.2389\).

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Key Concepts

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

Understanding Z-Scores
Z-scores are a way to describe how many standard deviations an element is from the mean of a set of data. In the context of the standard normal distribution, the mean is 0, and the standard deviation is 1. This makes it particularly easy to calculate probabilities and compare different data points because you are always working within the same scale.
Z-scores are calculated using the formula:
  • \[ z = \frac{(X - \mu)}{\sigma} \]This formula takes a raw score (X), subtracts the mean (\(\mu\)) of the distribution, and divides by the standard deviation (\(\sigma\)).
    • Essentially, this standardizes the data point X.
    • If a Z-score is positive, the data point is above the mean. If it's negative, it falls below the mean.
    • Z-scores allow us to easily calculate probabilities using the Z-table. This is because the table contains pre-calculated areas under the standard normal curve.
    • These scores are pivotal in probability problems, like the ones found in standard distributions.
Probability Calculations in Standard Normal Distribution
Probabilities in standard normal distribution refer to finding the area under the curve for a given Z-score. In other words, it is computing the likelihood that a random variable falls within a given range.Whenever you're asked to compute a probability like \[ P(a \leq z \leq b) \]this involves determining the area under the standard normal distribution curve from \( z = a \) to \( z = b \).
  • If the probability calculations involve only one side of the curve (i.e., \( P(z \leq c) \) or \( P(z > c) \)), the Z-score and area can be found directly from the Z-table.
  • Probability calculations like this can be referred to areas under the curve, ranging from the cumulative probability from the mean to a positive/negative Z-score. This means you often need to subtract one related probability from another, like \( P(a \leq z \leq b) = P(z \leq b) - P(z \leq a) \).
  • These calculations enable us to determine the likelihood of certain outcomes or intervals within a normally distributed variable.
Using the Z-Table for Probability Calculations
The Z-table is fundamental in finding probabilities associated with Z-scores. This table lists common Z-scores and their respective cumulative probabilities, which represent the total area under the curve from the left up to that score.
  • To use the Z-table effectively:
    - Locate the row that matches the first digit and first decimal place of your Z-score.
    - Find the column that matches the second decimal place of the Z-score.
    - The intersection of this row and column gives you the cumulative probability.
  • Remember, this probability is from the mean up to your desired Z-score.
  • For probabilities that are greater than a Z-score (e.g., \( P(z > x) \)), you subtract the table value from 1, since it represents the complementary probability.
  • Familiarity with reading the Z-table allows efficient solving of standard normal distribution problems by simply cross-referencing values.
In practice, the Z-table can be a powerful tool to assess probabilities for any standard normal distribution problem, allowing you to quickly handle questions, such as calculating the probability of very specific ranges.

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Most popular questions from this chapter

According to Advertising Age, the average base salary for women working as copywriters in advertising firms is higher than the average base salary for men. The average base salary for women is \(\$ 67,000\) and the average base salary for men is \(\$ 65,500\) (Working Woman, July/August 2000 ). Assume salaries are normally distributed and that the standard deviation is \(\$ 7000\) for both men and women. a. What is the probability of a woman receiving a salary in excess of \(\$ 75,000 ?\) b. What is the probability of a man receiving a salary in excess of \(\$ 75,000 ?\) c. What is the probability of a woman receiving a salary below \(\$ 50,000 ?\) d. How much would a woman have to make to have a higher salary than \(99 \%\) of her male counterparts?

In January \(2003,\) the American worker spent an average of 77 hours logged on to the Internet while at work (CNBC, March 15,2003 ). Assume the population mean is 77 hours, the times are normally distributed, and that the standard deviation is 20 hours. a. What is the probability that in January 2003 a randomly selected worker spent fewer than 50 hours logged on to the Internet? b. What percentage of workers spent more than 100 hours in January 2003 logged on to the Internet? c. \(A\) person is classified as a heavy user if he or she is in the upper \(20 \%\) of usage. In January \(2003,\) how many hours did a worker have to be logged on to the Internet to be considered a heavy user?

Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(-1.98 \leq z \leq .49)\) b. \(P(.52 \leq z \leq 1.22)\) c. \(\quad P(-1.75 \leq z \leq-1.04)\)

The unemployment rate is \(5.8 \%\) (Bureau of Labor Statistics, http: \(/ /\) www.bls.gov, April 3 , 2003 ). Suppose that 100 employable people are selected randomly. a. What is the expected number who are unemployed? b. What are the variance and standard deviation of the number who are unemployed? c. What is the probability that exactly six are unemployed? d. What is the probability that at least four are unemployed?

The average travel time to work for New York City residents is 36.5 minutes (Time Almanac, 2001 ). a. Assume the exponential probability distribution is applicable and show the probability density function for the travel time to work for a typical New Yorker. b. What is the probability it will take a typical New Yorker between 20 and 40 minutes to travel to work? c. What is the probability it will take a typical New Yorker more than 40 minutes to travel to work?
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