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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. Without assuming population normality, 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. 0.0062 b. Approximately 0

Step by step solution

01

Understand the Problem

We have a population with a mean hardness of \( \mu = 50 \) and a standard deviation of \( \sigma = 1.2 \). We need to find the probability that the sample mean is at least 51 for a given sample size in two scenarios: a smaller sample of 9 pins and a larger sample of 40 pins.
02

Sample Mean Distribution (a)

For part (a), the sample mean follows a normal distribution because the population is normal. For a sample of 9 pins, the distribution of the sample mean \( \bar{x} \) has a mean of \( \mu_{\bar{x}} = \mu = 50 \) and a standard deviation of \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{9}} = 0.4 \).
03

Calculate Z-score (a)

We compute the Z-score for the sample mean using the formula \( Z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \). Substituting the values, we get \( Z = \frac{51 - 50}{0.4} = 2.5 \).
04

Find Probability from Z-score (a)

Using the Z-score table, we find the probability \( P(Z \geq 2.5) \). The probability is \( 1 - P(Z \leq 2.5) \approx 1 - 0.9938 = 0.0062 \).
05

Sample Mean Distribution (b)

For part (b), due to the central limit theorem, the sample mean distribution can be approximated as normal for a sample size of 40. The mean of the sample mean is \( \mu_{\bar{x}} = 50 \) and the standard deviation is \( \sigma_{\bar{x}} = \frac{1.2}{\sqrt{40}} \approx 0.19 \).
06

Calculate Z-score (b)

Similarly, calculate the Z-score for part (b): \( Z = \frac{51 - 50}{0.19} \approx 5.26 \).
07

Find Probability from Z-score (b)

Using the Z-score table for \( Z = 5.26 \), we find that \( P(Z \geq 5.26) \) is practically 0 as the probability for very high Z-scores tends to near zero.

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Key Concepts

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

Normal Distribution
In statistics, a normal distribution is a common continuous probability distribution. It is often referred to as a bell curve due to its bell-shaped appearance. A normal distribution is symmetrical, meaning it has equal data points on the left and right of its central peak, which is also its mean.
Key characteristics of a normal distribution include:
  • The mean, median, and mode are equal and located at the center of the distribution.
  • It is defined by two parameters: the mean ( \(\mu\)) and the standard deviation ( \(\sigma\)).
  • Approximately 68% of the data lies within one standard deviation from the mean, 95% within two, and 99.7% within three.
In the exercise, the hardness of pins has a normal distribution with a mean of 50, which implies that the data clusters closely around this mean value.
Central Limit Theorem
The Central Limit Theorem (CLT) is a key principle in statistics that allows us to use normal distribution approximations for sample means, regardless of the distribution of the original data, given a sufficiently large sample size. In simpler terms, it states that the distribution of sample means will be approximately normally distributed if the sample size is large enough.
In question (b) of the exercise, even though we do not assume the population follows a normal distribution, we can still proceed using the Central Limit Theorem because the sample size of 40 is large, hence the sampling distribution of the mean will be approximately normal. This allows us to calculate probabilities using normal distribution assumptions even when the original data distribution might not be normal.
Z-Score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. Values can be positive or negative, indicating whether they are above or below the mean, respectively.
To calculate the Z-score, use the formula:\[Z = \frac{X - \mu}{\sigma}\]In the exercise, the Z-score is used to determine how far the sample mean of 51 is from the population mean of 50 in terms of standard deviations. This helps in finding the probability that the sample mean is at least 51. For example, in the smaller sample of 9 pins, the Z-score is calculated to be 2.5.
Sample Mean
The sample mean ( \(\bar{x}\)) is the average of a set of observations or data points from a larger population. It is a statistic used to estimate the population mean when it is not feasible to measure the entire population.
Calculating the sample mean involves adding all sample values together and dividing by the number of sample points:\[\bar{x} = \frac{\sum X_i}{n}\]where \(X_i\) are the sample values and \(n\) is the number of samples.
In our exercise, the sample mean is considered for samples of 9 and 40 pins to find the probability that the average hardness is at least 51. The sample mean varies depending on sample size due to the distribution properties.
Standard Deviation
Standard deviation ( \(\sigma\)) is a measure of the amount of variation or dispersion in a set of values. It is a widely used measure in statistics to quantify the degree of variation or spread in a data set.
A lower standard deviation indicates that the data points are close to the mean, while a higher standard deviation indicates that the data points are spread out over a larger range of values.
  • It is calculated as the square root of the variance.
  • The variance is the average of the squared differences from the mean.
In the given exercise, the population standard deviation is 1.2. We use this to calculate the standard deviation of the sample mean (\( \sigma_{\bar{x}}\)), which helps in calculating the Z-score and determining the probability of observing a sample mean of at least 51.

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Most popular questions from this chapter

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