/*! 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 57 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 200 and a standard deviation of 25 . a. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is approximately . 6330 . b. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is approximately \(.05\). c. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is. 8051 . d. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is \(.0150\). e. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is \(.4525\) 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 \(.4800\) and the value of \(x\) is greater than \(\mu\).

Short Answer

Expert verified
a) \(x\) approximately equals 208.75, b) \(x\) approximately equals 241.13, c) \(x\) approximately equals 179, d) \(x\) approximately equals 145.75, e) \(x\) approximately equals 198.75, f) \(x\) approximately equals 251.25

Step by step solution

01

Calculate Z-Score for Problem a

The Z-score associated with the probability of .6330 is approximately 0.35. Now solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = 0.35\).
02

Calculate Z-Score for Problem b

The given probability .05 represents the right tail, so 1 - .05 = .95 gives the area to the left. The Z-score associated with a probability of .95 is approximately 1.645. Solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = 1.645\).
03

Calculate Z-Score for Problem c

For this problem, the given probability .8051 represents the right tail, so 1 - .8051 = .1949 gives the area to the left. Calculating the Z-score associated with .1949, we get -0.88. Solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = -0.88\).
04

Calculate Z-Score for Problem d

The Z-score associated with a probability of .0150 is approximately -2.17. Solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = -2.17\).
05

Calculate Z-Score for Problem e

The given area between \(\mu\) and \(x\) is .4525 and \(x < \mu\). Since the area to the left of the mean is 0.5, the remaining area to \(x\) would be 0.5 - .4525 = .0475. The associated Z-score is approximately -0.05. Solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = -0.05\).
06

Calculate Z-Score for Problem f

Here, the area between \(\mu\) and \(x\) is .4800 and \(x > \mu\). So, the total area to the left of \(x\) is 0.5 + 0.4800 = 0.9800. The Z-score associated with a probability of .9800 is approximately 2.05. Solve \(Z = \frac{x - \mu}{\sigma}\) for \(x\) where \(Z = 2.05\).

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Key Concepts

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

Z-score calculation
The Z-score is a numerical measurement that describes a value's relation to the mean of a group of values. To calculate a Z-score, you use the formula: \( Z = \frac{x - \mu}{\sigma} \), where:
  • \( x \) is the value in question.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.
A Z-score tells you how many standard deviations an element is from the mean. A Z-score of 0 indicates the value is exactly at the mean. Positive Z-scores represent values above the mean, while negative Z-scores represent values below the mean.
In the context of a normal distribution, Z-scores help us find probabilities and identify percentile ranks. This is useful for various tasks, such as determining what proportion of data falls below a certain point in a normal distribution. Understanding how to convert areas into Z-scores and then into actual values of \( x \) is crucial when working with normal distributions.
Probability Density Function
A probability density function (PDF) is a function that describes the likelihood of a random variable to take on a given value. For a continuous random variable, the PDF assigns a probability to all outcomes in the space of the random variable, helping us understand the distribution.
The normal distribution, often known as the Gaussian distribution, is one of the most frequently used probability distributions in statistics due to its excellent naturally occurring properties. Its PDF is described mathematically by the equation: \[ f(x | \mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \]
  • \( \mu \) is the mean.
  • \( \sigma \) is the standard deviation.
This bell-shaped curve shows that data near the mean are more frequent in occurrence than data far from the mean.
The PDF not only helps in understanding the data but is instrumental for other statistical calculations and analysis, especially when working with the normal distribution and the normal curve behaviors.
Area under the curve
The area under the curve of a probability density function, in the context of the normal distribution, represents the probability of a random variable falling within a particular range. The total area under the standard normal distribution curve is always equal to 1, which represents the total probability of all possible outcomes.
When dealing with normal distributions, finding the area to the left or right of a certain point can give us the cumulative probability up to that point. This is why it is so useful to convert percentages or probabilities into Z-scores.
To find specific values, like in our exercise with questions a to f, you would:
  • Use a Z-score table or calculator to find the cumulative probability.
  • Apply the Z-score formula to determine the value of \( x \).
These tools are crucial to solve practical problems involving normal distributions, such as determining threshold values in quality control or marking grades on a curve. Grasping the concept of area under the curve not only aids in understanding distributions but also in making informed predictions grounded in statistical analysis.

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Most popular questions from this chapter

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\). a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

A company that has a large number of supermarket grocery stores claims that customers who pay by personal check spend an average of \(\$ 87\) on groceries at these stores with a standard deviation of \(\$ 22\). Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than \(\$ 114\) on groceries. b. What percentage of customers paying by check spend between \(\$ 40\) and \(\$ 60\) on groceries? c. What percentage of customers paying by check spend between \(\$ 70\) and \(\$ 105\) ? d. Is it possible for a customer paying by check to spend more than \(\$ 185\) ? Explain.

For a binomial probability distribution, \(n=80\) and \(p=.50 .\) Let \(x\) be the number of successes in 80 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \geq 42)\) using the normal approximation. c. Find \(P(41 \leq x \leq 48)\) using the normal approximation.

At Jen and Perry Ice Cream Company, the machine that fills 1 -pound cartons of Top Flavor ice cream is set to dispense 16 ounces of ice cream into every carton. However, some cartons contain slightly less than and some contain slightly more than 16 ounces of ice cream. The amounts of ice cream in all such cartons have a normal distribution with a mean of 16 ounces and a standard deviation of 18 ounce. a. Find the probability that a randomly selected carton contains \(16.20\) to \(16.50\) ounces of ice cream. b. What percentage of such cartons contain less than \(15.70\) ounces of ice cream? c. Is it possible for a carton to contain less than \(15.20\) ounces of ice cream? Explain.

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