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Find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the right of (a) \(z=-3.49\) (b) \(z=-0.55\) (c) \(z=2.23\) (d) \(z=3.45\)

Short Answer

Expert verified
a) 0.9998 b) 0.7088 c) 0.0129 d) 0.0003

Step by step solution

01

Understand the Standard Normal Curve

The standard normal curve is a bell-shaped curve that is symmetric about the mean, which is zero. The total area under the standard normal curve is 1. The area to the right of a z-score represents the probability that a value is greater than that z-score.
02

Determine the area to the right of z = -3.49

To find the area to the right of z = -3.49, use the standard normal distribution table or a calculator. The area to the left of z = -3.49 is approximately 0.0002. Hence, the area to the right is 1 - 0.0002 = 0.9998.
03

Determine the area to the right of z = -0.55

Find the area to the left of z = -0.55. Using the standard normal distribution table, the area to the left of z = -0.55 is approximately 0.2912. Therefore, the area to the right is 1 - 0.2912 = 0.7088.
04

Determine the area to the right of z = 2.23

For z = 2.23, look up the area to the left of this z-score in the table, which is approximately 0.9871. Therefore, the area to the right is 1 - 0.9871 = 0.0129.
05

Determine the area to the right of z = 3.45

For z = 3.45, find the area to the left of this z-score, which is approximately 0.9997. Hence, the area to the right is 1 - 0.9997 = 0.0003.

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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 tells us how many standard deviations a data point is from the mean of a distribution. The mean of a standard normal distribution is 0, and its standard deviation is 1. If a data point has a z-score of -3.49, it means it is 3.49 standard deviations below the mean. Conversely, a z-score of 2.23 indicates the data point is 2.23 standard deviations above the mean.
To calculate a z-score, use the formula: \[ z = \frac{(X - μ)}{σ} \] where \(X\) is the data point, \(μ\) is the mean, and \(σ\) is the standard deviation. Z-scores help compare data points from different normal distributions. The standard normal table, or z-table, lists the cumulative probabilities for each z-score, which are essential for finding areas under the curve.
Area Under the Curve in Standard Normal Distribution
The area under the standard normal curve represents probabilities. Because the curve is symmetric around the mean (0), the total area under the curve sums to 1. The areas to the left or right of a z-score give probabilities of an event occurring less than or greater than that z-score. For example, to find the area to the right of \(z = -3.49\), look up the cumulative probability for \(z = -3.49\) (0.0002) and subtract it from 1: \(1 - 0.0002 = 0.9998\).
Similarly, the area to the right of \(z = -0.55\) is calculated by finding the cumulative probability for \(z = -0.55\) (0.2912) and subtracting from 1: \(1 - 0.2912 = 0.7088\). These calculations help in understanding how likely an event is to occur in a standard normal distribution.
Calculating Probability Using Z-scores
Probability in the context of a standard normal distribution tells us how often a particular outcome is expected relative to the entire set of possible outcomes. In this setup:
  • A z-score translates a specific data point to a location on the standard normal curve
  • The area under the curve gives the cumulative probability
  • By using z-scores and the standard normal table, you can find the probability of a random variable falling within a certain range.
For instance, to find the likelihood of a value being greater than \(z = 2.23\), you refer to the cumulative probability: \[1 - P(Z < 2.23) = 1 - 0.9871 = 0.0129\] This indicates that there's a 1.29% chance of getting a value greater than 2.23 standard deviations above the mean.
Similar steps can be done for \(z = 3.45\), showing that only about 0.03% of data points lie to the right of \(3.45\). This helps in probabilistic decision making and inferential statistics.

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Most popular questions from this chapter

In a recent poll, the Gallup Organization found that \(45 \%\) of adult Americans believe that the overall state of moral values in the United States is poor. Suppose a survey of a random sample of 500 adult Americans is conducted in which they are asked to disclose their feelings on the overall state of moral values in the United States. Use the normal approximation to the binomial to approximate the probability that (a) exactly 250 of those surveyed feel the state of morals is poor. (b) no more than 220 of those surveyed feel the state of morals is poor. (c) more than 250 of those surveyed feel the state of morals is poor. (d) between 220 and 250 , inclusive, believe the state of morals is poor. (e) at least 260 adult Americans believe the overall state of moral values is poor. Would you find this result unusual? Why?

Assume that the random variable \(X\) is normally distributed, with mean \(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. \(P(56 \leq X<66)\)

The lives of refrigerators are normally distributed with mean \(\mu=14\) years and standard deviation \(\sigma=2.5\) years Source: Based on information from Consumer Reports (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of refrigerators that last for more than 17 years. (c) Suppose the area under the normal curve to the right of \(x=17\) is 0.1151 . Provide two interpretations of this result.

Steel rods are manufactured with a mean length of 25 centimeters \((\mathrm{cm}) .\) Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of \(0.07 \mathrm{~cm} .\) (a) What proportion of rods has a length less than \(24.9 \mathrm{~cm} ?\) (b) Any rods that are shorter than \(24.85 \mathrm{~cm}\) or longer than \(25.15 \mathrm{~cm}\) are discarded. What proportion of rods will be discarded? (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between \(24.9 \mathrm{~cm}\) and \(25.1 \mathrm{~cm} ?\)

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. O-Ring Thickness A random sample of O-rings was obtained, and the wall thickness (in inches) of each was recorded. $$\begin{array}{lllll} \hline 0.276 & 0.274 & 0.275 & 0.274 & 0.277 \\ \hline 0.273 & 0.276 & 0.276 & 0.279 & 0.274 \\ \hline 0.273 & 0.277 & 0.275 & 0.277 & 0.277 \\ \hline 0.276 & 0.277 & 0.278 & 0.275 & 0.276 \\ \hline \end{array}$$

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