/*! 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 71 Consider the population of all 1... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the population of all 1 -gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=\) \(5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture. Use the normal distribution model to calculate the following probabilities: a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(\quad P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

Short Answer

Expert verified
The probabilities are as follows: a. \(P(x<5.0) = 0.5\) b. \(P(x<5.4) = 0.9772\), c. \(P(x \leq 5.4) = 0.9772\), d. \(P(4.64.5)=0.9998\), f. \(P(x>4.0)=1\).

Step by step solution

01

Understand Z-Score

The Z-score normalized the values for the distribution model. The formula to compute Z-Score is \(Z = \frac{(X - \mu)}{\sigma}\) where X is the given value, \(\mu\) is the mean and \(\sigma\) is the standard deviation.
02

Calculate Z-scores for each part

Calculate Z scores for each part of the exercise as given. For example, for part a, \(Z_a = \frac{(5.0 - 5)}{0.2} = 0\), for part b, \(Z_b = \frac{(5.4 - 5)}{0.2} = 2.0\), and so on.
03

Determine the probability from Z-table

After calculating the Z-scores, we need to find the corresponding probabilities from the Z-table. For example, for \(Z_a = 0, P(x < 5.0) = P(Z < 0) = 0.5\) because the Z-table tells us that a Z-score of 0.00 corresponds to a probability of 0.500. Similarly, determine the probabilities for each part.
04

Compute the remaining probabilities

For the part d that have range, compute the probabilities separately and then subtract the smaller one from the larger one. For the part e and f that uses 'greater than', use the formula \(P(Z>k) = 1-P(Z \leq k)\) to find the probabilities.

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Key Concepts

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

Z-Score
The concept of a Z-score is pivotal in the realm of the normal distribution. It provides a way to understand how far away a data point is from the mean. A Z-score tells us the number of standard deviations a particular value is from the mean. The formula for calculating the Z-score is:\[ Z = \frac{(X - \mu)}{\sigma} \]Here:
  • \( X \) is the value in the dataset.
  • \( \mu \) represents the mean of the distribution.
  • \( \sigma \) is the standard deviation.
For example, if you have a mean of 5 ml and a standard deviation of 0.2 ml, and you're looking at the value 5.4 ml, the Z-score would be calculated as \( \frac{(5.4 - 5)}{0.2} = 2.0 \). This tells you that 5.4 ml is two standard deviations above the mean. Z-scores can be positive or negative, indicating whether the value is above or below the mean, respectively.
Probability Calculation
Probability calculation in the context of normal distribution often involves determining the likelihood of a value falling within a certain range. This is done using the Z-score and a Z-table. Once we have the Z-score, we can find the probability of the occurrence of that score by consulting the Z-table.To calculate the probability for a specific case (e.g., \( P(x < 5.0) \)), we first find the Z-score and then look up this score in the Z-table. If the Z-score for 5.0 ml is 0, we look up 0 in the table which gives a probability of 0.5. This means there's a 50% chance that a value is less than 5.0 ml in this normally distributed dataset. For ranges, such as \( P(4.6 < x < 5.2) \), calculate Z-scores for 4.6 and 5.2, find their probabilities, and subtract the smaller probability from the larger. This difference provides the probability that a value falls within this range. For 'greater than' scenarios, like \( P(x > 4.5) \), use the complement rule: calculate \( 1 - P(x \leq 4.5) \) to determine the probability of the value exceeding 4.5.
Standard Deviation
Standard deviation is a crucial concept in understanding how spread out the values in a dataset are. In the context of the normal distribution, it helps measure the average distance of each data point from the mean. A smaller standard deviation indicates that the values tend to be closer to the mean. Conversely, a larger standard deviation indicates more variability in the data, with values spanning a wider range. In our paint example, the standard deviation is 0.2 ml. This implies the typical difference between any given amount of red dye and the mean is 0.2 ml. It's important because it allows us to understand the distribution's spread and, together with the mean, to calculate other statistical measures like Z-scores and probabilities. When working with standard deviation, always remember it gives context to the mean, allowing for a more comprehensive picture of data distribution.
Z-Table
The Z-table, sometimes known as the standard normal table, is an essential tool for converting Z-scores into probabilities. The table lists Z-scores and their corresponding probabilities for a standard normal distribution. To use the Z-table, locate the Z-score in the table. The intersection of the row and column where your Z-score sits gives you the probability of a random variable being less than or equal to that score. For instance, if you have a Z-score of 1.5, the Z-table will provide the cumulative probability from the left up to that Z-score. For scores not listed explicitly in the table, interpolate by using the values of the closest scores. Additionally, probabilities for negative Z-scores can be found using the symmetry of the normal distribution. You simply take the positive equivalent value of the Z-score. The Z-table is indispensable for converting Z-scores to probabilities, enabling us to determine the likelihood of different outcomes within the normal distribution.

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Most popular questions from this chapter

Suppose that \(20 \%\) of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeowners are selected at random; let \(x\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability \((.2)(.8)(.2)(.2)\) and associated \(x\) value of \(3 .\) There are 15 other outcomes.) b. What is the most likely value of \(x\) ? c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

Determine each of the following areas under the standard normal \((z)\) curve: a. To the left of \(-1.28\) b. To the right of \(1.28\) c. Between \(-1\) and 2 d. To the right of 0 e. To the right of \(-5\) f. Between \(-1.6\) and \(2.5\) g. To the left of \(0.23\)

Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=\) 500 and \(p=.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (b)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities computed in Part (c), write a sentence interpreting the probability.

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Give examples of two different assignments of probabilities such that the resulting distributions have the same mean but quite different standard deviations.

Exercise \(7.9\) introduced the following probability distribution for \(y=\) the number of broken eggs in a carton: $$ \begin{array}{cccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & .01 \end{array} $$ a. Calculate and interpret \(\mu_{y} .\) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y}^{?}\) Does this surprise you? c. Why doesn't \(\mu_{y}=(0+1+2+3+4) / 5=2.0\). Explain.

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