/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Find the \(z\) value described a... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the \(z\) value described and sketch the area described.Find \(z\) such that \(97.5 \%\) of the standard normal curve lies to the left of \(z\).

Short Answer

Expert verified
The z-score is 1.96.

Step by step solution

01

Understanding the Task

We need to find a value of z, often called the z-score, such that 97.5% of the standard normal curve (which has a mean of 0 and a standard deviation of 1) is to the left of this z-score.
02

Locating the Appropriate z-Score

To locate the z-score, we can use the standard normal distribution table (also known as the Z-table) or a calculator. We look for an entry that comes closest to 0.975, which represents 97.5%.
03

Finding z in the Z-table

By looking at the Z-table, we find that a z-score of 1.96 corresponds to a cumulative probability of approximately 0.975, which means that 97.5% of the distribution lies to the left of z = 1.96.
04

Visualizing the Area

On a standard normal curve, highlight the area to the left of the found z-score, which is 1.96. This shaded area represents 97.5% of the total area under the curve, confirming that z = 1.96 is correct.

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Key Concepts

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

Z-score
A z-score is a statistical measure that tells us how many standard deviations a data point is from the mean of a data set. It's a way to quantify the position of a score in a distribution. The formula for calculating the z-score of a given value is:\[ z = \frac{{X - \mu}}{{\sigma}} \]where:
  • \(X\) is the data point
  • \(\mu\) is the mean of the data set
  • \(\sigma\) is the standard deviation of the data set
In the context of a standard normal distribution, which has a mean of 0 and a standard deviation of 1, each z-score corresponds to a point on the curve.
Since the standard normal distribution is symmetrical and bell-shaped, a z-score of 0 is precisely at the mean, and positive or negative scores indicate distances to the right or left of the mean, respectively.
The importance of the z-score lies in its ability to help us understand the location and significance of a specific value in relation to the overall distribution. In this example, finding a z-score where 97.5% of the curve lies to the left involves searching for a particular z-score that satisfies this condition.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a certain value. In the context of the standard normal distribution, it is the probability that a random data point lies within the left-hand side of the curve up to a specific z-score.
Essentially, when we discuss cumulative probability, we're summing up probabilities from the beginning of the range up to the designated value. Mathematically, if you're looking for a cumulative probability of 0.975, it means that 97.5% of the data points are below that particular z-score.
This is crucial because cumulative probability allows us to understand how data points are distributed across intervals on a statistical curve. By analyzing cumulative probabilities, we get valuable insights into how much data lies below (or above) a certain threshold in the total data set.
Z-table
The Z-table, also known as the standard normal table, is an essential statistical tool for dealing with the standard normal distribution. It provides the cumulative probabilities of z-scores—essentially, it tells you the percentage of data points falling to the left of a particular z-score on a standard normal curve.
To use the Z-table, locate the desired probability value on the table. Each cell in the Z-table represents a cumulative probability. Our task is to find the probability closest to 0.975 in the table, which indicates the proportion of the distribution that is to the left of the target z-score.
Once the closest value is found, trace back to the corresponding z-score value on the table. In our example, the z-score corresponding to 0.975 in the Z-table is 1.96. This means that 97.5% of data points on a standard normal distribution lie to the left of z = 1.96.
The Z-table is a vital reference for statisticians because it facilitates the interpretation and visualization of data in a standardized manner, making complex calculations accessible and understandable.

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Most popular questions from this chapter

Watches Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 28 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 5 months, and the distribution of lifetimes is normal. (a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production should the company expect to replace? (b) Inverse Normal Distribution If Accrotime does not want to make refunds on more than \(12 \%\) of the watches it makes, how long should the guarantee period be (to the nearest month)?

Find the \(z\) value described and sketch the area described.Find \(z\) such that \(82 \%\) of the standard normal curve lies to the right of \(z\).

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).

(a) If we have a distribution of \(x\) values that is more or less mound-shaped and somewhat symmetric, what is the sample size needed to claim that the distribution of sample means \(\bar{x}\) from random samples of that size is approximately normal? (b) If the original distribution of \(x\) values is known to be normal, do we need to make any restriction about sample size in order to claim that the distribution of sample means \(\bar{x}\) taken from random samples of a given size is normal?

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(-2.18 \leq z \leq-0.42)$$

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