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Let \(z\) denote a variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38\) ) h. \(P(z<4.98)\)

Short Answer

Expert verified
The calculated probabilities are the solutions needed for each part. The answers will vary based on the reference tables used. Prospective students must perform these steps individually to obtain their specific probability values

Step by step solution

01

Interpret the Inequalities

The symbol '<' denotes that we are interested in the probability that z is less than the given value (excluding the value itself), while '≤' includes the value. Similarly, '>' denotes the value to be more and '>=' means more or equal. Use the standard normal distribution table or a calculator to get the associated probabilities.
02

Determine Probabilities

Given that 'z' is a variable from a standard normal distribution, use a standard normal distribution table or calculator to find the area under the curve to the left of the given z-score for 'less than' or 'less than or equal to' equations (a, b, c, h). For 'greater than' or 'greater than or equal to' statements (f, g), you need to find the area to the right of the given z-score, which is \(1- P(z \leq value)\). For ranges (d, e), find the area between the two z-scores.
03

Calculate for All Cases

a. Given that z is a variable from a standard normal distribution, to find \(P(z<2.36)\), look up 2.36 in the standard normal distribution table. b. \(P(z \leq 2.36)\) would be equal to \(P(z<2.36)\) as the likelihood of z being exactly 2.36 is infinitely small in a continuous distribution. c. To find \(P(z<-1.23)\), again, use the table to get the associated probability. d. To find \(P(1.142)\), find \(P(z<2)\) from the table and subtract it from 1. g. \(P(z \geq-3.38)\) would be 1-\(P(z<-3.38)\). h. \(P(z<4.98)\) can be found directly from the table.
04

Cross Verification

Verify the calculated probabilities using a standard normal calculator if one is available to confirm the results obtained in step 3.

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Key Concepts

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

Understanding Z-Scores
A z-score represents the number of standard deviations a data point is from the mean of a distribution. In the realm of a standard normal distribution, this score is pivotal for interpreting the position of a value within this distribution. It is simply calculated by subtracting the mean from the value in question and dividing that by the standard deviation. Since the standard normal distribution has a mean of 0 and a standard deviation of 1, the z-score in this case is the actual value of interest.

For example, a z-score of 2.36 means that the observed value is 2.36 standard deviations above the mean. Conversely, a z-score of -1.23 indicates that the value is 1.23 standard deviations below the mean. Understanding the concept of z-scores is crucial for interpreting probability questions in statistics, as they allow us to standardize different data points for comparison.
Probability in the Standard Normal Distribution
Probability, in the context of a standard normal distribution, indicates the likelihood of a z-score occurring within a given range. This distribution is symmetric around the mean (which is zero for the standard normal distribution), and the probabilities for specific ranges of z-scores can be determined using the area under the distribution curve.

For instance, the probability that a randomly selected z-score is less than 2.36, denoted as \(P(z<2.36)\), corresponds to the area under the standard normal curve to the left of z=2.36. This is a cumulative probability, which means it includes all z-scores from negative infinity up to 2.36. When we talk about probabilities, we are usually looking for the cumulative area under the curve that aligns with our specific z-score scenario.
Utilizing the Normal Distribution Table
A normal distribution table, often referred to as a z-table, lists the cumulative probabilities associated with z-scores in a standard normal distribution. To find the probability for a specific z-score, one must locate the z-score in the table to determine the area to the left of that score.

The table typically provides the probabilities from the mean up to the z-value, due to the symmetry of the distribution. Thus, to find the cumulative probability for positive z-scores, you would use the table directly. However, for negative z-scores or probabilities to the right, you may need to perform additional calculations – such as subtracting the table value from 1 – to obtain the correct probability. Understanding how to read and interpret the z-table is an essential skill for students studying statistics.
Continuous Distribution Features
A continuous distribution, such as the standard normal distribution, represents a range of possible values for a continuous variable – values that can take on any number within a given range. Unlike discrete distributions that are based on countable outcomes, a continuous distribution is described by a density curve, and the probability of observing any single, exact value is zero. Instead, probabilities are associated with ranges of values.

In continuous distributions, calculations involving inequalities (such as \(P(z<2.36)\)) are understood to incorporate all the values up to but not including the specified value. The probability for a range of values (for example, \(P(1.14

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Most popular questions from this chapter

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean \(45 \mathrm{~min}\) and standard deviation \(5 \mathrm{~min}\). a. If \(50 \mathrm{~min}\) is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if we wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Let \(x\) denote the systolic blood pressure of an individual selected at random from a certain population. Suppose that the normal distribution with mean \(\mu=\) \(120 \mathrm{~mm} \mathrm{Hg}\) and standard deviation \(\sigma=10 \mathrm{~mm} \mathrm{Hg}\) is a reasonable model for describing the population distribution of \(x\). (The article "Oral Contraceptives, Pregnancy, and Blood Pressure," Journal of the American Medical Association [1972]: \(1507-1510\), reported on the results of a large study in which a sample histogram of blood pressures among women of similar ages was found to be well approximated by a normal curve.) a. Calculate \(P(110145)\), the probability that \(x\) is more than \(2.5\) standard deviations from its mean value.

Consider the variable \(x=\) earthquake insurance status for the population of homeowners in an earthquake-prone California county. This variable associates a category (insured or not insured) with each individual in the population. a. Construct a relative frequency bar chart that represents the population distribution for \(x\) for the case where \(60 \%\) of the county homeowners have earthquake insurance. b. If an individual is randomly selected from this population, what is the probability that the selected homeowner does not have earthquake insurance?

A certain basketball player makes \(70 \%\) of his free throws. Assume that results of successive free throws are independent of one another. At the end of a particular practice, the coach tells the player to begin shooting free throws and to stop only when he has made two consecutive shots. Let \(x\) denote the number of shots until the player can stop. Describe how you would carry out a simulation experiment to approximate the distribution of \(x\).

The light bulbs used to provide exterior lighting for a large office building have an average lifetime of \(700 \mathrm{hr}\). If the distribution of the variable \(x=\) length of bulb life can be modeled as a normal distribution with a standard deviation of \(50 \mathrm{hr}\), how often should all the bulbs be replaced so that only \(20 \%\) of the bulbs will have already burned out?

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