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For a normal random variable \(x\) with mean \(\mu\) and standard deviation \(\sigma\) specified in the exercises. \(\mu=1.2\) and \(\sigma=.15 .\) Find \(P(1.00

Short Answer

Expert verified
Answer: The probability, P(1.00 < x < 1.10), is approximately 0.1607.

Step by step solution

01

Write down the given information.

We are given the normal random variable \(x\) with mean \(\mu = 1.2\) and standard deviation \(\sigma = 0.15\). The probability we want to find is \(P(1.00 < x < 1.10)\).
02

Standardize the given values.

We need to convert the values \(1.00\) and \(1.10\) to their corresponding z-scores using the formula: \(z = \frac{x - \mu}{\sigma}\). For \(x = 1.00\), the z-score (\(z_1\)) is: \(z_1 = \frac{1.00 - 1.2}{0.15} = -1.3333\) For \(x = 1.10\), the z-score (\(z_2\)) is: \(z_2 = \frac{1.10 - 1.2}{0.15} = -0.6667\)
03

Find the corresponding probabilities from the standard normal table.

Referring to the standard normal table, look up the probabilities corresponding to \(z_1\) and \(z_2\): \(P(z < -1.3333) = 0.0918\) \(P(z < -0.6667) = 0.2525\)
04

Calculate the probability \(P(1.00 < x < 1.10)\).

We can find the probability \(P(1.00 < x < 1.10)\) by subtracting the probabilities from Step 3: \(P(1.00 < x < 1.10) = P(z < -0.6667) - P(z < -1.3333) = 0.2525 - 0.0918 = 0.1607\) So, the probability \(P(1.00 < x < 1.10)\) is approximately \(0.1607\).

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of dispersion in statistics, informing us about the variability of data points around the mean (average). It is an essential tool for understanding how spread out a distribution of values is. In the context of a normal random variable, the standard deviation, denoted as \(\text{\(\sigma\)}\), quantifies the degree to which individual observations in a dataset deviate from the dataset's mean, \(\mu\).

The higher the standard deviation, the more the data points are spread out from the mean, and vice versa. If you're visualizing this, imagine a bell curve; a smaller standard deviation means the bell is narrow and tall, indicating that most data points are close to the mean. A larger standard deviation results in a wider and flatter bell, showing that the data points are more spread out. When applied to problem-solving, calculating standard deviation helps us understand the probability of a random variable falling within a certain range, which brings us to using z-scores for further analysis.
Z-Score Calculation
Z-score calculation is a statistic that measures the number of standard deviations a data point is from the mean of the distribution. It is a critical concept in probability and statistics, allowing us to standardize an entire distribution and simplify the process of finding probabilities.

To calculate the z-score, you use the formula \(z = \frac{x - \mu}{\sigma}\), where \(x\) is an observation from your data, \(\mu\) is the mean of the data, and \(\sigma\) is the standard deviation. By standardizing the scores (thus creating z-scores), we can compare different sets of data that may have different means and standard deviations on a standard scale. This is especially useful for normal distributions. For example, in the exercise mentioned, converting the values into z-scores allows us to use the standard normal table to find the associated probabilities, moving us towards our final answer.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution, characterized by a mean of 0 and a standard deviation of 1. It is the graphical representation of the z-scores, which follows a symmetrical, bell-shaped curve called the 'normal curve'. This is where the concept of z-scores plays a significant role; any normal distribution can be translated into the standard normal distribution using z-scores.

A key element in working with the standard normal distribution is the standard normal table (or z-table). This table lists the cumulative probabilities associated with z-scores, allowing us to determine the probability of a z-score being less than or equal to a particular value. In our exercise, we converted the range of values into their corresponding z-scores and then referred to the standard normal table to calculate the desired probability. The result is a demonstration of how the standard normal distribution is used to find probabilities within any normal distribution.

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Most popular questions from this chapter

The diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 10 centimeters and a standard deviation of 3.75 centimeters. a. What proportion of the trees will have diameters between 7.5 and 12.5 centimeters? b. What proportion of the trees will have diameters less than 7.5 centimeters? c. Your Christmas tree stand will expand to a diameter of 15 centimeters. What proportion of the trees will not fit in your Christmas tree stand?

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A purchaser of electric relays buys from two suppliers, \(A\) and \(B\). Supplier A supplies two of every three relays used by the company. If 75 relays are selected at random from those in use by the company, find the probability that at most 48 of these relays come from supplier A. Assume that the company uses a large number of relays.

The length of life (in days) of an alkaline battery has an exponential distribution with an average life of 1 year, so that \(\lambda=1 / 365\). a. What is the probability that an alkaline battery will fail before 180 days? b. What is the probability that an alkaline battery will last beyond 1 year? c. If a device requires two batteries, what is the probability that both batteries last beyond 1 year?

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