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Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(-0.73 \leq z \leq 3.12)$$

Short Answer

Expert verified
The probability is 0.7664.

Step by step solution

01

Understanding the Problem

We are asked to find the probability that a standard normal random variable \( z \) is between -0.73 and 3.12. This probability corresponds to the area under the standard normal curve between these two \( z \)-values.
02

Using the Standard Normal Distribution Table

Consult a standard normal distribution (Z) table to find the cumulative probability of \( z = -0.73 \) and \( z = 3.12 \). The table provides the probability that a standard normal variable is less than a given \( z \)-value.
03

Finding Cumulative Probabilities

From the Z table, find \( P(z \leq -0.73) \) and \( P(z \leq 3.12) \). Typically, you take the value from the row and column that meet the \( z \) value. \( P(z \leq -0.73) = 0.2327 \) and \( P(z \leq 3.12) = 0.9991 \).
04

Calculating the Probability

Subtract the cumulative probability of the lower \( z \)-value from the higher \( z \)-value to find the probability in the specified range. \( P(-0.73 \leq z \leq 3.12) = P(z \leq 3.12) - P(z \leq -0.73) = 0.9991 - 0.2327 = 0.7664 \).
05

Shading the Area Under the Curve

On the standard normal distribution curve, shade the area between \( z = -0.73 \) and \( z = 3.12 \). This area represents the probability found in the previous step.

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special type of normal distribution. It has a mean (c) of 0 and a standard deviation (cc) of 1. This creates a bell-shaped curve that is symmetrical around the mean. The standard normal distribution is used to standardize normal distributions, making it easier to compare different normal distributions or transform data.

In practical terms, any normal distribution can be converted into a standard normal distribution using a transformation formula. This is usually done to simplify calculations or comparisons:
  • For any given data point, you calculate its Z-score using the formula: \[Z = \frac{(X - \mu)}{\sigma}\]where \(X\) is the data point, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation.
  • The resulting Z-score tells you how many standard deviations a particular value is from the mean.
Z-table
A Z-table, also known as a standard normal distribution table, is used to find cumulative probabilities associated with standard normal distribution Z-scores. These tables help us understand the probability that a random variable falls below a certain value in a standard normal distribution.

The Z-table is organized by rows and columns, each corresponding to whole and fractional parts of Z-scores:
  • The rows typically represent the leading Z score digit and one decimal place (e.g., -0.7).
  • The columns add an extra hundredth place (e.g., 0.03), making it easy to find exact Z-scores such as -0.73.
By finding the intersection of a given row and column in the Z-table, you can easily locate the cumulative probability for any specific Z-score.
Cumulative Probability
Cumulative probability refers to the probability that a random variable takes on a value less than or equal to a specific value. In the context of the standard normal distribution and Z-table, it provides the total area under the curve to the left of a specific Z-score.

For example, if you want to find the cumulative probability for a Z-score of -0.73, you look it up in the Z-table and find the associated value, such as 0.2327. This value means that there's a 23.27% chance that a variable will fall below a Z-score of -0.73.

The cumulative probability for a higher Z-score of 3.12 from the Z-table might be 0.9991, indicating nearly all potential outcomes fall below this point. To compute the probability between two Z-scores, subtract the cumulative probability of the lower score from that of the higher score. This gives the portion of outcomes that lie between them.
Area Under the Curve
The area under the curve in a standard normal distribution graphically represents the cumulative probability. Each part of the curve corresponds to probabilities of specific Z-value ranges.

For example, to find the probability that a variable lies between two Z-scores, you calculate the area under the curve that falls between these scores. This is visualized as a shaded region on the standard normal distribution curve.
  • Calculate the cumulative probability for the upper Z-score.
  • Calculate the cumulative probability for the lower Z-score.
  • Subtract the cumulative probability of the lower Z-score from the upper one to find the area (probability) between them.
In our example, this area between Z = -0.73 and Z = 3.12 represents the probability of 0.7664. This shaded area is the focus when visualizing or interpreting results.

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Most popular questions from this chapter

Let \(x\) be a random variable that represents checkout time (time spent in the actual checkout process in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the \(x\) distribution is about \(\mu=2.7\) minutes, with standard deviation \(\sigma=0.6\) minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of \(x\) values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let \(x_{i}(\text { for } i=1,2,3, \ldots, 30)\) represent the checkout time for each customer. For example, \(x_{1}\) is the checkout time for the first customer, \(x_{2}\) is the checkout time for the second customer, and so forth. Each \(x_{i}\) has mean \(\mu=2.7\) minutes and standard deviation \(\sigma=0.6\) minute. Let \(w=x_{1}+x_{2}+\cdots+x_{30}\). Explain why the problem is asking us to compute the probability that \(w\) is less than \(90 .\) (b) Use a little algebra and explain why \(w<90\) is mathematically equivalent to \(w / 30<3 .\) since \(w\) is the total of the \(30 x\) values, then \(w / 30=\bar{x} .\) Therefore, the statement \(\bar{x}<3\) is equivalent to the statement \(w<90 .\) From this we conclude that the probabilities \(P(\bar{x}<3)\) and \(P(w<90)\) are equal. (c) What does the central limit theorem say about the probability distribution of \(\bar{x} ?\) Is it approximately normal? What are the mean and standard deviation of the \(\bar{x}\) distribution? (d) Use the result of part (c) to compute \(P(\bar{x}<3) .\) What does this result tell you about \(P(w<90) ?\)

The manager of Motel 11 has 316 rooms in Palo Alto, California. From observation over a long period of time, she knows that on an average night, 268 rooms will be rented. The long-term standard deviation is 12 rooms. This distribution is approximately mound-shaped and symmetric. (a) For 10 consecutive nights, the following numbers of rooms were rented each night: $$\begin{array}{l|cccccc} \hline \text { Night } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Number of rooms } & 234 & 258 & 265 & 271 & 283 & 267 \\ \hline & & & & & & \\ \hline \text { Night } & 7 & 8 & 9 & 10 & & \\ \hline \text { Number of rooms } & 290 & 286 & 263 & 240 & & \\ \hline \end{array}$$ Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Interpretation Looking at the control chart, would you say the number of rooms rented during this 10-night period has been unusually low? unusually high? about what you expected? Explain your answer. Identify all out-of-control signals by type (I, II, or III). Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Interpretation Looking at the control chart, would you say the number of rooms rented during this 10-night period has been unusually low? unusually high? about what you expected? Explain your answer. Identify all out-of-control signals by type (I, II, or III). (b) For another 10 consecutive nights, the following numbers of rooms were rented each night: $$\begin{array}{l|cccccc} \hline \text { Night } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Number of rooms } & 238 & 245 & 261 & 269 & 273 & 250 \\ \hline & & & & & & \\ \hline \text { Night } & 7 & 8 & 9 & 10 & & \\ \hline \text { Number of rooms } & 241 & 230 & 215 & 217 & & \\ \hline \end{array}$$ Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Would you say the room occupancy has been high? low? about what you expected? Explain your answer. Identify all out-of- control signals by type (I, II, or IIII).

Basic Computation: Normal Approximation to a Binomial Distribution Suppose we have a binomial experiment with \(n=40\) trials and probability of success \(p=0.85.\) (a) Is it appropriate to use a normal approximation to this binomial distribution? Why? (b) Compute \(\mu\) and \(\sigma\) of the approximating normal distribution. (c) Use a continuity correction factor to convert the statement \(r<30\) successes to a statement about the corresponding normal variable \(x .\) (d) Estimate \(P(r<30).\) (e) Interpretation Is it unusual for a binomial experiment with 40 trials and probability of success 0.85 to have fewer than 30 successes? Explain.

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. To the right of \(z=1.52\)

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \leq 0)$$

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