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Draw a figure of the standard normal curve showing: a. \(z(0.04)\) b. \(z(0.94)\)

Short Answer

Expert verified
The standard normal distribution is a symmetrical, bell-shaped curve focused around the mean value of 0. \(z(0.04)\) is a point on the left of mean (0) and \(z(0.94)\) is a point on the right of mean (0). The areas under the curve to the left of these z-scores are 0.04 and 0.94 respectively.

Step by step solution

01

Understanding the z-Score

The notation \(z(0.04)\) and \(z(0.94)\) represents a z-score. In context of the standard normal distribution, a z-score shows how many standard deviations a given data point is from the mean. It's negative if the data point is less than the mean and positive if it's greater than the mean. Since the mean of standard normal distribution is 0, the z-score also tells you the location of that data point in relation to the center of distribution or where it is on the x-axis. The values in parentheses, represent the area to the left of z-score under the curve. For example, \(z(0.04)\) means the area to the left of this point under curve is 0.04 or 4%.
02

Draw the Standard Normal Curve

Draw a bell-shaped curve. Label the x-axis in units of standard deviations from the mean, marking the mean (0) in the middle. The curve should be symmetrical about this mean.
03

Identify the Position of \(z(0.04)\)

The z-score corresponding to a cumulative probability of 0.04 falls on the left side of the mean because 0.04 is less than 0.50 (which is the center). Mark this approximate point as \(z(0.04)\) on the curve.
04

Identify the Position of \(z(0.94)\)

The z-score corresponding to a cumulative probability of 0.94 falls on the right side of the mean because 0.94 is more than 0.50 (which is the center). Mark this approximate point as \(z(0.94)\) on the curve.
05

Shading the Area

Shade the area under the curve to the left of \(z(0.04)\) and \(z(0.94)\). These shaded areas represent areas with probabilities 0.04 and 0.94 respectively.

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Key Concepts

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

z-score
The z-score is a key concept in statistics, vital for understanding the standard normal distribution. It indicates how many standard deviations an element is from the mean. If you imagine all data points perfectly distributed along a horizontal line, the mean sits at the center with a z-score of 0. Every step to the right or left represents one standard deviation, either positive or negative respectively.

In the original exercise, the notation like z(0.04) is essentially asking for the z-score that has 4% of the data to its left. This reveals its position relative to the mean. Understanding z-scores can help students interpret a variety of data sets in real-world scenarios, from test scores to heights, and temperatures. Knowing how to calculate and interpret a z-score transforms a raw data point into actionable insight.
normal curve
The normal curve, also known as the bell curve, is a graphical representation of the standard normal distribution. It's named for its bell-like shape, symmetrically centered around the mean. This curve is essential as it illustrates the distribution of scores or values.

The furthest points on each end represent the extremes and contain the least frequent occurrences. Most data clusters around the middle, or mean, and tapers off as it moves away. To draw it, you would mark the mean at the center, often labeled as zero in the case of a standardized distribution, and plot the frequency of occurrence vertically. In the textbook exercise, shading the area to the left of certain z-scores visualizes what proportion of data falls within those boundaries.
cumulative probability
Cumulative probability is the likelihood that a variable will take a value less than or equal to a particular level. On a normal curve, it's represented by the area under the curve to the left of a z-score. This area quantifies the 'accumulated' likelihood, up to the given point.

This is essential for interpreting data or making predictions. If, for instance, a student's z-score on a standardized test is found to be at z(0.94), this tells us that they've scored better than 94% of the test-takers. Why? Because the cumulative probability up to their z-score covers 94% of the total area under the curve. By understanding cumulative probabilities, students can not only evaluate results but also set thresholds for success in probabilistic terms.

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Most popular questions from this chapter

Based on a survey conducted by Greenfield Online, 25 to 34-year-olds spend the most each week on fast food. The average weekly amount of \(\$ 44\) was reported in a May 2009 USA Today Snapshot. Assuming that weekly fast food expenditures are normally distributed with a standard deviation of \(\$ 14.50,\) what is the probability that a 25- to 34-year-old will spend: a. less than \(\$ 25\) a week on fast food? b. between \(\$ 30\) and \(\$ 50\) a week on fast food? c. more than \(\$ 75\) a week on fast food?

Find the area under the standard normal curve between \(z=-2.75\) and \(z=-1.28, P(-2.75

Use a computer or calculator to find the probability that one randomly selected value of \(x\) from a normal distribution, with mean 584.2 and standard deviation 37.3 will have a value a. less than 525. b. between 525 and 590. c. of at least 590. d. Verify the result using Table 3. e. Explain any differences you may find.

Find the area under the normal curve that lies between the following pairs of \(z\) -values: a. \(\quad z=-1.20\) to \(z=-0.22\) b. \(\quad z=-1.75\) to \(z=-1.54\) c. \(\quad z=1.30\) to \(z=2.58\) d. \(\quad z=0.35\) to \(z=3.50\)

A company that manufactures rivets used by commercial aircraft manufacturers knows that the shearing strength of (force required to break) its rivets is of major concern. They believe the shearing strength of their rivets is normally distributed, with a mean of 925 pounds and a standard deviation of 18 pounds. a. If they are correct, what percentage of their rivets has a shearing strength greater than 900 pounds? b. What is the upper bound for the shearing strength of the weakest \(1 \%\) of the rivets? c. If one rivet is randomly selected from all of the rivets, what is the probability that it will require a force of at least 920 pounds to break it? d. Using the probability found in part c, what is the probability, rounded to the nearest tenth, that 3 rivets in a random sample of 10 will break at a force less than 920 pounds?

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