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a. Find the probability that \(z\) is greater than \(-.75 .\) b. Find the probability that \(z\) is less than 1.35 .

Short Answer

Expert verified
Answer: (a) The probability that z is greater than -0.75 is 0.2266. (b) The probability that z is less than 1.35 is 0.9115.

Step by step solution

01

Finding the probability that z is greater than -0.75

In this step, we will find the probability that the z-score is greater than -0.75. Since the standard normal distribution is symmetrical about the mean, the area to the right of -0.75 is equal to the area to the left of 0.75. First, we need to check the standard normal distribution table for the probability corresponding to z = 0.75. The table value for z = 0.75 is 0.7734. This means that the area to the left of z = 0.75 is 0.7734. Since the total area under the curve is equal to 1, we can calculate the area to the right of z = -0.75 as follows: P(z > -0.75) = 1 - P(z < 0.75) = 1 - 0.7734 = 0.2266 So, the probability that z is greater than -0.75 is 0.2266.
02

Finding the probability that z is less than 1.35

In this step, we will find the probability that the z-score is less than 1.35. We need to check the standard normal distribution table for the probability corresponding to z = 1.35, which is given by the area to the left of z = 1.35. The table value for z = 1.35 is 0.9115. This means that the area to the left of z = 1.35 is 0.9115. So, the probability that z is less than 1.35 is 0.9115. Now, we have the answers to both tasks: a. The probability that z is greater than -0.75 is 0.2266. b. The probability that z is less than 1.35 is 0.9115.

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Key Concepts

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

Standard Normal Distribution
In probability theory, the concept of the standard normal distribution is fundamental. It is a type of normal distribution that is fully characterized by having a mean (\( \mu \)) of 0 and a standard deviation (\( \sigma \)) of 1. This distribution is denoted as\( N(0,1) \). The standard normal distribution is represented by a bell-shaped curve that is symmetrical around its mean.

Key features of the standard normal distribution include:
  • Symmetrical shape: The curve is identical on both sides of the mean.
  • Total area under the curve: It sums up to 1, covering the probability of all possible outcomes.
  • 68-95-99.7 rule: About 68% of values lie within one standard deviation from the mean, 95% within two, and 99.7% within three.
Transforming any normal distribution to a standard normal distribution simplifies calculations, especially when utilizing z-scores and probability tables.
Z-Score
A z-score is an essential statistical measurement that expresses the number of standard deviations a data point is from the mean of a distribution. It is calculated using the formula:\[z = \frac{(X - \mu)}{\sigma}\]Here, \( X \) represents the value for which the z-score is being calculated, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.

Z-scores help in evaluating how extreme a point is within the crux of a distribution. If a z-score is:
  • Positive: the value is above the mean.
  • Negative: the value is below the mean.
  • Zero: the value is equal to the mean.
By converting raw scores to z-scores, we can easily compare data from different distributions and determine probabilities using the standard normal distribution.
Probability Table
A probability, or standard normal distribution, table is a handy mathematical tool used to find the probability of a z-score in a standard normal distribution. It provides the cumulative probability that a normally distributed random variable is less than or equal to a given z-score.

Using the probability table involves the following:
  • Identifying the z-score for which you want to find the probability.
  • Locating the row corresponding to the integer and first decimal place of the z-score.
  • Finding the column corresponding to the second decimal place of the z-score.
For example, if you are looking for the probability that a z-score is less than 1.35, you would find 0.9115 in the table. The table assumes values to the left of the curve. For values to the right, you subtract the table value from 1, accommodating the entire distribution's total area of 1.

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Most popular questions from this chapter

A manufacturing plant uses 3000 electric light bulbs whose life spans are normally distributed, with mean and standard deviation equal to 500 and 50 hours, respectively. In order to minimize the number of bulbs that burn out during operating hours, all the bulbs are replaced after a given period of operation. How often should the bulbs be replaced if we wish no more than \(1 \%\) of the bulbs to burn out between replacement periods?

A normal random variable \(x\) has mean \(\mu=1.20\) and standard deviation \(\sigma=.15 .\) Find the probabilities of these \(x\) -values: a. \(1.001.38\) c. \(1.35

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Let \(x\) be a binomial random variable with \(n=\) 36 and \(p=.54 .\) Use the normal approximation to find: a. \(P(x \leq 25)\) b. \(P(15 \leq x \leq 20)\) c. \(P(x>30)\)

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