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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of \(1.2\). a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51 ? b. What is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 ?

Short Answer

Expert verified
a. Probability ≈ 0.0062; b. Probability ≈ 0.

Step by step solution

01

Understanding the Problem

We are given the mean and standard deviation of a population distribution, and we need to find the probability of the sample mean exceeding a certain value for two samples of different sizes.
02

Define the Parameters

The population mean (\( \mu \)) is 50, and the population standard deviation (\( \sigma \)) is 1.2. We are evaluating two cases with sample sizes (\( n \)) of 9 and 40 pins.
03

Calculate the Standard Error (n=9)

The standard error of the sample mean when \( n = 9 \) is given by:\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{9}} = \frac{1.2}{3} = 0.4 \]
04

Find the Z-Score for Sample Mean 51 (n=9)

The Z-score for a sample mean of 51 is calculated as:\[ Z = \frac{\bar{x} - \mu}{SE} = \frac{51 - 50}{0.4} = \frac{1}{0.4} = 2.5 \]
05

Determine Probability using Z-Score (n=9)

Using the standard normal distribution table, we find the probability that \( Z > 2.5 \). This corresponds to \( P(Z > 2.5) \approx 0.0062 \).
06

Calculate the Standard Error (n=40)

For a sample size of 40, the standard error is:\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{40}} \approx \frac{1.2}{6.3246} \approx 0.19 \]
07

Find the Z-Score for Sample Mean 51 (n=40)

The Z-score for a sample mean of 51 with \( n = 40 \) is:\[ Z = \frac{\bar{x} - \mu}{SE} = \frac{51 - 50}{0.19} \approx \frac{1}{0.19} \approx 5.26 \]
08

Determine Probability using Z-Score (n=40)

Again using the standard normal distribution table, we estimate the probability that \( Z > 5.26 \), which is extremely small and essentially \( 0 \).

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Key Concepts

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

Probability Calculations
Probability calculations help us understand the likelihood of certain events occurring. In the context of a normal distribution, this involves using the sample mean and other parameters to determine how likely it is for the sample mean to exceed a particular threshold.

Imagine you have a large pile of pins, and you want to know the probability that a small group of them will have a hardness value above a certain point. You won't always have the exact result due to natural variations in the sample. That's where probability helps by giving an estimate based on the available data.
  • Probability calculations provide an estimate, not a certainty.
  • They rely on the characteristics of the normal distribution and the given parameters.
Using these calculations, you can predict events like whether your sample mean will be higher or lower than expected.
Z-Score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. In the context of normal distribution, it indicates how many standard deviations an element is from the mean.

To calculate a Z-score, follow these simple steps:
  • Subtract the population mean from the raw score (in our case, the sample mean).
  • Divide the result by the standard error of the mean.
The formula for the Z-score is:
\[ Z = \frac{\bar{x} - \mu}{SE} \] Where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the population mean.
  • SE is the standard error.
Understanding Z-scores allows you to determine the probability of observing a sample mean that is a certain number of standard deviations away from the population mean.
Standard Error
The standard error measures the variability of the sample mean estimate from the population mean. It's essential for probability calculations because it provides a sense of how much the sample mean can be expected to vary from the true population mean.

The formula for calculating the standard error (SE) is: \[ SE = \frac{\sigma}{\sqrt{n}} \] Where:
  • \(\sigma\) is the population standard deviation.
  • \(n\) is the sample size.
The larger the sample size, the smaller the standard error, meaning the sample mean is more likely to be closer to the population mean. Conversely, a smaller sample size results in a larger standard error.

A small standard error indicates precise sample mean estimates, while a large one suggests more variability.
Sample Mean
The sample mean is the average of all data points in a sample and is used to make inferences about the population mean. It is a critical component in statistical analysis because it serves as an unbiased estimator of the population mean.

When working with samples, you calculate the mean by adding up all the observed values and dividing by the number of observations:
\[ \bar{x} = \frac{\sum{x_i}}{n} \] Where:
  • \(x_i\) is each individual observation.
  • \(n\) is the sample size.
The sample mean not only helps in probability calculations but also in determining the Z-score, which further aids in understanding the probability of various outcomes.

It is important to choose a sample large enough to represent the population accurately to gain meaningful insights from the sample mean.

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Most popular questions from this chapter

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\lambda=50\). What is the approximate probability that a. Between 35 and 70 tickets are given out on a particular day? [Hint: When \(\lambda\) is large, a Poisson rv has approximately a normal distribution.] b. The total number of tickets given out during a 5-day week is between 225 and 275 ?

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5\) min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

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Let \(X_{1}, X_{2}, \ldots, X_{100}\) denote the actual net weights of 100 randomly selected 50 -lb bags of fertilizer. a. If the expected weight of each bag is 50 and the variance is 1 , calculate \(P(49.9 \leq \bar{X} \leq 50.1\) ) (approximately) using the CLT. b. If the expected weight is \(49.8 \mathrm{lb}\) rather than \(50 \mathrm{lb}\) so that on average bags are underfilled, calculate \(P(49.9 \leq \bar{X} \leq 50.1)\).

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