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A simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\). (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

Short Answer

Expert verified
a) Normal distribution, mean=64, standard error=3. b) 0.3192 c) 0.0571 d) 0.6549

Step by step solution

01

- Describe the sampling distribution of \(\bar{x}\)

The sampling distribution of the sample mean \(\bar{x}\) can be described using the Central Limit Theorem, which states that if we have a sufficiently large sample size, the sampling distribution of \(\bar{x}\) will be approximately normally distributed. In this case, the population mean \(\text{μ} = 64\) and the population standard deviation \(\text{σ} = 18\). The mean of the sampling distribution \(\text{μ}_{\bar{x}}\) is equal to the population mean, \(\text{μ}_{\bar{x}} = 64\), and the standard deviation of the sampling distribution (standard error) \(\text{σ}_{\bar{x}}\) is given by \(\frac{\text{σ}}{\text{√n}} = \frac{18}{\text{√36}} = 3\).
02

- Calculate \(P(\bar{x}

To find \(P(\bar{x}<62.6)\), convert the sample mean to a z-score using the formula \(z = \frac{x - μ_{\bar{x}}}{σ_{\bar{x}}}\): \[ z = \frac{62.6 - 64}{3} = \frac{-1.4}{3} = -0.467\]. Then, use a z-table or standard normal distribution calculator to find the probability corresponding to z = -0.47. \(P(\bar{x}<62.6)\) is approximately 0.3192.
03

- Calculate \(P(\bar{x} \geq\ 68.7)\)

To find \(P(\bar{x} \geq\ 68.7)\), convert the sample mean to a z-score: \[z = \frac{68.7 - 64}{3} = \frac{4.7}{3} = 1.567\]. Using a z-table or a standard normal distribution calculator, find the probability corresponding to z = 1.57 and subtract it from 1. The probability of z being 1.57 is approximately 0.9429, so \(P(\bar{x} \geq\ 68.7) = 1 - 0.9429 = 0.0571\).
04

- Calculate \(P(59.8

To find \(P(59.8<\bar{x}<65.9)\), convert both sample means to z-scores: \[z_1 = \frac{59.8 - 64}{3} = \frac{-4.2}{3} = -1.4\] and \[z_2 = \frac{65.9 - 64}{3} = \frac{1.9}{3} = 0.633\]. Use the z-table to find the probabilities corresponding to z = -1.4 and z = 0.633. \(P(z < -1.4)\) is approximately 0.0808, and \(P(z < 0.63)\) is approximately 0.7357. Thus, \(P(59.8<\bar{x}<65.9) = 0.7357 - 0.0808 = 0.6549\).

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Key Concepts

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

Sampling Distribution
The sampling distribution of \( \bar{x} \) refers to the probability distribution of the sample mean from a set of samples taken from the same population. According to the Central Limit Theorem, the sampling distribution of \( \bar{x} \) tends to be normal if the sample size is sufficiently large, even if the population distribution is not normal. For our case, where \( n = 36 \), the theorem applies, and we can state that the sampling distribution of \( \bar{x} \) is approximately normal.

The mean of the sampling distribution (\( \text{μ}_{\bar{x}} \)) is equal to the population mean (\( \text{μ}=64 \)), and the standard deviation of the sampling distribution (also known as the standard error) is calculated using the formula: \(\text{σ}_{\bar{x}} = \frac{\text{σ}}{\text{√n}} \) which in our example simplifies to \( \frac{18}{6} = 3 \). Therefore, \( \text{μ}_{\bar{x}} = 64 \) and \(\text{σ}_{\bar{x}} = 3 \).
Z-Score
A Z-score quantifies the number of standard deviations a data point is from the mean of the sampling distribution. It helps in standardizing different data points so we can compare them.

For instance, to find the probability \( P(\bar{x}<62.6) \), we compute the Z-score using the formula: \( z = \frac{x - μ_{\bar{x}}}{σ_{\bar{x}}} \). Substituting into the equation, \( z = \frac{62.6 - 64}{3} = -0.467 \). This Z-score tells us how far away 62.6 is from the average sample mean in standard deviation units.
Standard Error
The standard error (SE) measures an estimate of the standard deviation of the sample mean's sampling distribution. It provides an understanding of how much variability we can expect in the sample mean from sample to sample.

From our example, with \( n = 36 \) and a population standard deviation of \( \text{σ} = 18 \), the SE is calculated as \( \frac{\text{σ}}{\text{√n}} = \frac{18}{6} = 3 \). The smaller the SE, the more accurate our sample mean is as an estimate of the population mean.

Probability Calculations
Probability calculations allow us to find the likelihood of a certain sample mean occurring within the sampling distribution.

For instance, to find \( P(\bar{x} \text{ < } 62.6) \), we compute the Z-score and then find the corresponding probability using a Z-table. Given \( z = -0.467 \), we find the probability of \( P(z < -0.467) \) which is approximately 0.3192.

Similarly, for \( P(\bar{x} \text{ ≥ } 68.7) \), the process involves computing \( z = 1.567 \), finding the area to the left from Z-tables and subtracting it from 1 to get the result: \( P(\bar{x} \text{ ≥ } 68.7) = 0.0571 \).
Population Mean
The population mean (\( \text{μ} \)) is the average of all values in the population. It is a fixed value and doesn't change. In our example, the population mean is given as 64.

In sampling, we use the population mean to compare with sample means. The goal is often to understand how sample results relate to the overall population.

When computing probabilities or Z-scores, we use this population mean as our benchmark value.
Standard Deviation
The standard deviation (\text{σ}) measures how spread out the values in a population are. It gives insights into data variability.

In our case, the population standard deviation is 18. When we take samples, we often calculate the standard error (SE), which is the standard deviation of the sampling distribution. For our given sample size, we use \( \text{σ}_{\bar{x}} = \frac{\text{σ}}{\text{√n}} = \frac{18}{6} = 3 \).

Standard deviation helps in understanding the dispersion of data and is a key component in calculating Z-scores and probabilities in a sampling distribution.

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Most popular questions from this chapter

Suppose you want to study the number of hours of sleep you get each evening. To do so, you look at the calendar and randomly select 10 days out of the next 300 days and record the number of hours you sleep. (a) Explain why number of hours of sleep in a night by you is a random variable. (b) Is the random variable "number of hours of sleep in a night" quantitative or qualitative? (c) After you obtain your ten nights of data, you compute the mean number of hours of sleep. Is this a statistic or a parameter? Why? (d) Is the mean number of hours computed in part (c) a random variable? Why? If it is a random variable, what is the source of variation?

Marriage Obsolete? According to a study done by the Pew Research Center, \(39 \%\) of adult Americans believe that marriage is now obsolete. (a) Suppose a random sample of 500 adult Americans is asked whether marriage is obsolete. Describe the sampling distribution of \(\hat{p}\), the proportion of adult Americans who believe marriage is obsolete. (b) What is the probability that in a random sample of 500 adult Americans less than \(38 \%\) believe that marriage is obsolete? (c) What is the probability that in a random sample of 500 adult Americans between \(40 \%\) and \(45 \%\) believe that marriage is obsolete? (d) Would it be unusual for a random sample of 500 adult Americans to result in 210 or more who believe marriage is obsolete?

Without doing any computation, decide which has a higher probability, assuming each sample is from a population that is normally distributed with \(\mu=100\) and \(\sigma=15 .\) Explain your reasoning. (a) \(P(90 \leq \bar{x} \leq 110)\) for a random sample of size \(n=10\). (b) \(P(90 \leq \bar{x} \leq 110)\) for a random sample of size \(n=20\).

A simple random sample of size \(n=1000\) is obtained from a population whose size is \(N=1,000,000\) and whose population proportion with a specified characteristic is \(p=0.35\). (a) Describe the sampling distribution of \(\hat{p}\). (b) What is the probability of obtaining \(x=390\) or more individuals with the characteristic? (c) What is the probability of obtaining \(x=320\) or fewer individuals with the characteristic?

What happens to the standard deviation of \(\hat{p}\) as the sample size increases? If the sample size is increased by a factor of 4 what happens to the standard deviation of \(\hat{p} ?\)

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