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Chapter 7: Problem 9

Suppose \(x\) has a distribution with \(\mu=15\) and \(\sigma=14\). (a) If a random sample of size \(n=49\) is drawn, find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(15 \leq \bar{x} \leq 17)\). (b) If a random sample of size \(n=64\) is drawn, find \(\mu_{\bar{x}}, \sigma_{\bar{x}}\), and \(P(15 \leq \bar{x} \leq 17)\). (c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)

Short Answer

Expert verified
The probability in part (b) is higher because a larger sample size reduces \(\sigma_{\bar{x}}\), increasing the probability within the specified interval.

Step by step solution

01

Understand the problem

We are given a distribution with a mean \( \mu = 15 \) and standard deviation \( \sigma = 14 \). We need to evaluate the sample distribution for samples of size 49 and 64, find their means (\( \mu_{\bar{x}} \)) and standard deviations (\( \sigma_{\bar{x}} \)), and calculate the probability that the sample mean \( \bar{x} \) is between 15 and 17.
02

Calculate \( \mu_{\bar{x}} \) for Part (a)

For a sample mean, \( \mu_{\bar{x}} = \mu \). Therefore, \( \mu_{\bar{x}} = 15 \) for both parts (a) and (b) because it is equal to the population mean.
03

Calculate \( \sigma_{\bar{x}} \) for Part (a)

For a sample size \( n = 49 \), the standard deviation of the sample mean \( \sigma_{\bar{x}} \) is calculated using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \). Thus, \( \sigma_{\bar{x}} = \frac{14}{\sqrt{49}} = 2 \).
04

Calculate Probability for Part (a)

We want \( P(15 \leq \bar{x} \leq 17) \). This probability can be found by converting to a standard normal distribution: \( P\left( \frac{15-15}{2} \leq Z \leq \frac{17-15}{2} \right) \). Thus, \( P(0 \leq Z \leq 1) \), which is approximately 0.3413.
05

Calculate \( \mu_{\bar{x}} \) for Part (b)

Since the sample mean equals the population mean, \( \mu_{\bar{x}} = 15 \) remains the same as in Part (a).
06

Calculate \( \sigma_{\bar{x}} \) for Part (b)

For a sample size \( n = 64 \), \( \sigma_{\bar{x}} = \frac{14}{\sqrt{64}} = 1.75 \).
07

Calculate Probability for Part (b)

Again, we evaluate \( P(15 \leq \bar{x} \leq 17) \) by converting to a standard normal distribution: \( P\left( \frac{15-15}{1.75} \leq Z \leq \frac{17-15}{1.75} \right) \). Thus, \( P(0 \leq Z \leq 1.14) \) which is approximately 0.3790.
08

Compare Results of Parts (a) and (b)

With a larger sample size, the standard deviation \( \sigma_{\bar{x}} \) decreases, causing the range 15 to 17 to cover a larger area under the standard normal curve. Hence, the probability for part (b) is higher.

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Key Concepts

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

Sample Mean
When engaging with statistical data, especially in a practical setting, one of the crucial elements you'll encounter is the 'Sample Mean.' The sample mean, denoted as \( \mu_{\bar{x}} \), is essentially the average of all the individual data points in a sample. It represents the center point or balance of the dataset.
Understanding the sample mean is important because it gives us an estimate of the overall population mean \( \mu \).
  • For large datasets, the sample mean tends to be very close to the population mean due to the Central Limit Theorem.
  • It is calculated by summing all sample data points and dividing by the number of points (\( n \)).
  • In statistical exercises, whether you draw a sample of 49 or 64, like the one given, the sample mean \( \mu_{\bar{x}} \) will always equal the population mean \( \mu = 15 \).
Overall, the sample mean serves as a vital tool in statistics, helping us draw conclusions and make predictions about a population based on a smaller, manageable sample size.
Standard Deviation
Another key concept in statistics is the 'Standard Deviation,' which describes the spread or variability of a set of data points. In the context of sample distribution, the standard deviation of the sample mean, \( \sigma_{\bar{x}} \), is particularly meaningful.
This value indicates how much the sample means deviate from the actual population mean.
  • It is derived from the population standard deviation \( \sigma \) and the sample size \( n \) using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).
  • This formula shows that as the sample size increases, \( \sigma_{\bar{x}} \) decreases, indicating a tighter clustering of sample means around the population mean.
  • For our given exercise, the standard deviation becomes smaller when the sample size increases from 49 to 64 (\( 2 \to 1.75 \)), showing why the sample mean becomes a more accurate reflection of the population mean with larger samples.
A smaller standard deviation of the sample mean suggests less variation, providing greater confidence in the accuracy of the sample mean as an estimate of the population mean.
Probability Distribution
In statistics, one might often refer to the concept of 'Probability Distribution.' This concept fundamentally describes how probabilities are spread across different outcomes or intervals.
Probability distributions are used to represent real-world phenomena in a mathematical form, making predictions possible.
  • For normally distributed data, which our given problem hints at, there's a bell-shaped curve involved. Here, most of the data points cluster around the mean, with probabilities tapering off symmetrically on either side.
  • Our focus is on the probability that the sample mean \( \bar{x} \) falls within a certain range. In our scenario, between 15 and 17.
  • As the exercise shows, converting this to the standard normal distribution \( Z \), allows the use of statistical tables to find probabilities.For example, when \( P(0 \leq Z \leq 1) \) is calculated, it corresponds to roughly 34.13% of the data.
  • With a larger sample, this probability increases due to the smaller spread or standard deviation, meaning that sample means are more concentrated around the population mean.
Understanding probability distribution helps in predicting the likelihood of various outcomes, and offers insights into data trends and variations.

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Most popular questions from this chapter

Expand Your Knowledge: Totals Instead of Averages Let \(x\) be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express lane of a large grocery. Based on a consumer survey, the mean of the \(x\) distribution is about \(\mu=2.7\) minutes, with standard deviation \(\sigma\) \(=0.6\) minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of \(x\) values is more or less symmetrical and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps. (a) Let \(x_{i}\) (for \(\left.i=1,2,3, \ldots, 30\right)\) represent the checkout time for each customer. For example, \(x_{1}\) is the checkout time for the first customer, \(x_{2}\) is the checkout time for the second customer, and so forth. Each \(x_{i}\) has mean \(\mu=2.7\) minutes and standard deviation \(\sigma=0.6\) minute. Let \(w=x_{1}+x_{2}+\cdots+x_{30 .}\) Explain why the problem is asking us to compute the probability that \(w\) is less than 90 . (b) Use a little algebra and explain why \(w<90\) is mathematically equivalent to \(w / 30<3 .\) Since \(w\) is the total of the \(30 x\) values, then \(w / 30=\bar{x}\). Therefore, the statement \(\bar{x}<3\) is equivalent to the statement \(w<90\). From this we conclude that the probabilities \(P(\bar{x}<3)\) and \(P(w<90)\) are equal. (c) What does the central limit theorem say about the probability distribution of \(\bar{x}\) ? Is it approximately normal? What are the mean and standard deviation of the \(\bar{x}\) distribution? (d) Use the result of part \((c)\) to compute \(P(\bar{x}<3)\). What does this result tell you about \(P(w<90)\) ?

Describe how the variability of the \(\bar{x}\) distribution changes as the sample size increases.

Medical: White Blood Cells Let \(x\) be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that \(x\) has a distribution that is approximately normal, with mean \(\mu=7500\) and estimated standard deviation \(\sigma=1750\) (see reference in Problem 13). A test result of \(x<3500\) is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, \(x\) is less than 3500 ? (b) Suppose a doctor uses the average \(\bar{x}\) for two tests taken about a week apart. What can we say about the probability distribution of \(\bar{x} ?\) What is the probability of \(\bar{x}<3500 ?\) (c) Repeat part (b) for \(n=3\) tests taken a week apart. (d) Interpretation: Compare your answers to parts (a), (b), and (c). How did the probabilities change as \(n\) increased? If a person had \(\bar{x}<3500\) based on three tests. what conclusion would vou draw as a doctor or a nurse?

Medical Tests: Leukemia Healthy adult bone marrow contains about \(56.5 \%\) neutrophils (a particular type of white blood cell). However, if this level is significantly reduced, it may be an early indicator of leukemia (Reference: Diagnostic Tests with Nursing Implications, edited by S. Loeb, Springhouse). (a) In a laboratory biopsy, a field of \(n=50\) bone marrow cells are observed under a microscope. A special dye is inserted, which only the neutrophils absorb. Then the number \(r\) of neutrophils in the field is counted. Although the field size \(n=50\) is fixed, the number of neutrophils \(r\) is a random variable, so the proportion \(\hat{p}=r / n\) is also a random variable. Explain why \(\hat{p}\) can be approximated by a normal random variable. What are \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\) (b) Suppose Jan had a bone marrow biopsy and \(\hat{p}\) was observed to be \(0.53\). Assuming nothing is wrong (no leukemia), what is the probability of getting a biopsy result this low or lower? Compute \(P(\hat{p} \leq 0.53)\). (c) Suppose Meredith had a bone marrow biopsy and \(\hat{p}\) was observed to be 0.41. Assuming nothing is wrong (no leukemia), what is the probability of getting a biopsy result this low or lower? Compute \(P(\hat{p} \leq 0.41)\). (d) Interpretation: Based on the probability estimates in parts (b) and (c), which do you think is the more serious case, Jan or Meredith? Explain.

Vital Statistics: Heights of Men The heights of 18 -year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches (based on information from Statistical Abstract of the United States, 112 th Edition). (a) What is the probability that an 18 -year-old man selected at random is between 67 and 69 inches tall? (b) If a random sample of nine 18 -year-old men is selected, what is the probability that the mean height \(\bar{x}\) is between 67 and 69 inches? (c) Interpretation: Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?
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