/*! 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 39 Let \(x\) be a continuous random... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x\) be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of \(75 .\) a. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is \(.0250\). b. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is \(.9345\). c. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is approximately .0275. d. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is approximately \(.9600\). e. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4700\) and the value of \(x\) is less than \(\mu\). f. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4100\) and the value of \(x\) is greater than \(\mu\).

Short Answer

Expert verified
Value of \(x\) for a to f can be found by substituting values into \(x = \mu + z \cdot \sigma\) after finding the corresponding z-score from the standard normal distribution table or using software packages like R or Excel.

Step by step solution

01

Calculate x for Part a

Considering the area to the left, we need to find the z-score corresponding to an area of .0250 in standard normal distribution table. Let's say this z-score value is \(z\). Using the formula \(x = \mu + z \cdot \sigma\), where \(\mu=550\) is mean and \(\sigma=75\) is standard deviation, we can calculate \(x\).
02

Calculate x for Part b

Since the area to the right of x is given, we need to subtract it from 1 to get the area to the left. After this, we can follow the same steps as part a.
03

Calculate x for Part c

Follow the same strategy as part b, since we're looking for the area to the right again.
04

Calculate x for Part d

The approach is the same as for part a. The area to the left is given directly.
05

Calculate x for Part e

The area between \(\mu\) and \(x\) is given. Since \(x\) is less than \(\mu\), the area to the left of \(x\) can be calculated by subtracting the area between \(\mu\) and \(x\) from .5. Then use it to find x in a similar way to previous parts.
06

Calculate x for Part f

The area between \(\mu\) and \(x\) is given. Since \(x\) is greater than \(\mu\), the area to the left of \(x\) can be calculated by adding the area between \(\mu\) and \(x\) to .5. Then use it to find x in a similar way to previous parts.

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Key Concepts

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

Continuous Random Variable
A continuous random variable is a variable that can take on any value within a given range. It is characterized by its ability to assume infinitely many values between two endpoints. Unlike discrete random variables, which can only take specific, separate values, continuous random variables have a seamless flow and represent measurable quantities. Examples include height, weight, and time.
For continuous random variables, probabilities are expressed over intervals rather than at specific points. The likelihood of the variable occurring within a certain range is found as the area under the curve of its probability density function (PDF). This is crucial when dealing with normal distribution, where the probabilities correspond to areas under the bell-shaped curve. This encompasses values such as heights between 5.5 and 6 feet tall or test scores ranging from 70 to 85.
Understanding continuous random variables is key to applying the concepts of the normal distribution effectively.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or spread in a set of values. It tells us how much the individual data points of a data set deviate from the mean or average of the set. A small standard deviation means that most of the numbers are close to the mean, whereas a large standard deviation indicates that the numbers are spread out.
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. This is known as the Empirical Rule, and it's foundational for predictions and conclusions relating to data.
Calculating standard deviation involves determining the square root of the average of the squared deviations from the mean. This helps in understanding the variability and consistency of data, which is vital in assessing normal distribution scenarios.
Z-score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. It expresses how many standard deviations away a particular score is from the mean. The formula for the z-score is:
  • \( z = \frac{x - \mu}{\sigma} \)
where \( x \) is the value we're interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Z-scores allow us to standardize different normal distributions to compare values from different datasets or systems. For example, if a test score of 700 has a z-score of +1.5, this means the score is 1.5 standard deviations above the mean. Z-scores are particularly useful because they enable easier calculation of the probability associated with a score in a normal distribution.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It is an important concept because it serves as a reference point for calculating probabilities and understanding the behavior of normal distributions in general.
In this distribution, scores are often converted into z-scores to align any normal distribution with the standard one. The standard normal distribution allows us to use z-tables or computational tools to determine the probability of a variable falling within a certain range or exceeding a certain value.
Key characteristics include its symmetrical bell shape, where the area under its curve corresponds to probability. Understanding this standard form aids in solving problems involving any normal distribution by simplifying and standardizing comparisons.

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Most popular questions from this chapter

Find the following binomial probabilities using the normal approximation. a. \(n=140, \quad p=.45, \quad P(x=67)\) b. \(n=100, \quad p=.55, \quad P(52 \leq x \leq 60)\) c. \(n=90, \quad p=.42, \quad P(x \geq 40)\) d. \(n=104, \quad p=.75, \quad P(x \leq 72)\)

Jenn Bard, who lives in the San Francisco Bay area, commutes by car from home to work. She knows that it takes her an average of 28 minutes for this commute in the morning. However, due to the variability in the traffic situation every morning, the standard deviation of these commutes is 5 minutes. Suppose the population of her morning commute times has a normal distribution with a mean of 28 minutes and a standard deviation of 5 minutes. Jenn has to be at work by \(8: 30\) A.M. every morning. By what time must she leave home in the morning so that she is late for work at most \(1 \%\) of the time?

Alpha Corporation is considering two suppliers to secure the large amounts of steel rods that it uses. Company A produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of \(.15 \mathrm{~mm}\) and sells 10,000 rods for $$\$ 400$$ Company B produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of \(.12 \mathrm{~mm}\) and sells 10,000 rods for $$\$ 460$$ A rod is usable only if its diameter is between \(7.8 \mathrm{~mm}\) and \(8.2 \mathrm{~mm}\). Assume that the diameters of the rods produced by each company have a normal distribution. Which of the two companies should Alpha Corporation use as a supplier? Justify your answer with appropriate calculations.

Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of \(2.5\) seconds. What percentage of his times between pitches are a. longer than 39 seconds? b. between 29 and 34 seconds?

A machine at Keats Corporation fills 64 -ounce detergent jugs. The machine can be adjusted to pour, on average, any amount of detergent into these jugs. However, the machine does not pour exactly the same amount of detergent into each jug; it varies from jug to jug. It is known that the net amount of detergent poured into each jug has a normal distribution with a standard deviation of \(.35\) ounce. The quality control inspector wants to adjust the machine such that at least \(95 \%\) of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be?

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